Codeforces Round #410 (Div. 2)(A,字符串,水坑，B,暴力枚举，C,思维题，D,区间贪心)

Mike has a string s consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.

A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.

Input

The first and single line contains string s (1 ≤ |s| ≤ 15).

Output

Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.

Examples

Input

abccaa

Output

YES

Input

abbcca

Output

NO

Input

abcda

Output

YES
题目链接：http://codeforces.com/contest/798/problem/A
题意:
改变一个字符，让这组字符串变成回文序列，能的话输出YES，不能输出NO
分析：
这道题有点坑，应该是卡了三组数据，aa输出NO，a输出YES，aba也输出YES，估计过了这三组，就可以AC了吧
这题我的做法是分奇偶性，判断条件s[i]==s[len-i-1]是否成立，成立的话ans++，然后判断奇数项长度ans==(len)/2-1||ans==len/2是否成立，成立输出YES，否则输出NO
偶数项长度只需判断ans==(len)/2-1是否成立，成立为YES，否则为NO
下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     char s[20];
 6     while(cin>>s)
 7     {
 8         int len=strlen(s);
 9         for(int i=0;i<len;i++)
10         {
11             for(int j='a';j<='z';j++)
12             {
13                 if(s[i]==j)
14                     continue;
15             }
16         }
17         int ans=0;
18         if(len%2!=0)
19         {
20             for(int i=0;i<(len)/2;i++)
21             {
22                 if(s[i]==s[len-i-1])
23                     ans++;
24             }
25             if(ans==(len)/2-1||ans==len/2)
26                 cout<<"YES"<<endl;
27             else cout<<"NO"<<endl;
28 
29         }
30         else if(len%2==0)
31         {
32         for(int i=0;i<(len)/2;i++)
33         {
34                 if(s[i]==s[len-i-1])
35                     ans++;
36         }
37             if(ans==(len)/2-1)
38                 cout<<"YES"<<endl;
39             else cout<<"NO"<<endl;
40         }
41     }
42     return 0;
43 }

Mike has n strings s₁, s₂, ..., s_n each consisting of lowercase English letters. In one move he can choose a string s_i, erase the first character and append it to the end of the string. For example, if he has the string "coolmike", in one move he can transform it into the string "oolmikec".

Now Mike asks himself: what is minimal number of moves that he needs to do in order to make all the strings equal?

Input

The first line contains integer n (1 ≤ n ≤ 50) — the number of strings.

This is followed by n lines which contain a string each. The i-th line corresponding to string s_i. Lengths of strings are equal. Lengths of each string is positive and don't exceed 50.

Output

Print the minimal number of moves Mike needs in order to make all the strings equal or print - 1 if there is no solution.

Examples

Input

4 
xzzwo 
zwoxz 
zzwox 
xzzwo

Output

Input

2 
molzv 
lzvmo

Output

Input

3 
kc 
kc 
kc

Output

Input

3 
aa 
aa 
ab

Output

-1

Note

In the first sample testcase the optimal scenario is to perform operations in such a way as to transform all strings into "zwoxz".

题目链接：http://codeforces.com/contest/798/problem/B

题意:

枚举一个字符串为标准，求答案，找最小

分析：

调用一个substr函数，具体函数详解请参看http://www.cnblogs.com/ECJTUACM-873284962/p/6763801.html

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 void change(string &s)
 4 {
 5     s=s.substr(1)+s.substr(0,1);//字符串截取函数,这里的目的就是一个一个换字符，找最小次数
 6 }
 7 int main()
 8 {
 9     int n;
10     while(cin>>n)
11     {
12         string s[n];
13         for(int i=0;i<n;i++)
14             cin>>s[i];
15         int ans=1e+8;
16         int sum;
17         for(int i=0;i<n;i++)
18         {
19             sum=0;
20             for(int j=0;j<n;j++)
21             {
22                 string t=s[j];
23                 while(t!=s[i])
24                 {
25                     change(t);
26                     sum++;
27                     if(sum>3000)
28                     {
29                         ans=-1;
30                         break;
31                     }
32                 }
33             }
34             ans=min(ans,sum);
35         }
36         cout<<ans<<endl;
37     }
38     return 0;
39 }

Mike has a sequence A = [a₁, a₂, ..., a_n] of length n. He considers the sequence B = [b₁, b₂, ..., b_n] beautiful if the gcd of all its elements is bigger than 1, i.e. $Codeforces Round #410 (Div. 2)(A,字符串,水坑，B,暴力枚举，C,思维题，D,区间贪心)$ .

Mike wants to change his sequence in order to make it beautiful. In one move he can choose an index i (1 ≤ i < n), delete numbers a_i, a_i + 1 and put numbers a_i - a_i + 1, a_i + a_i + 1 in their place instead, in this order. He wants perform as few operations as possible. Find the minimal number of operations to make sequence A beautiful if it's possible, or tell him that it is impossible to do so.

$Codeforces Round #410 (Div. 2)(A,字符串,水坑，B,暴力枚举，C,思维题，D,区间贪心)$ is the biggest non-negative number d such that d divides b_i for every i (1 ≤ i ≤ n).

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — length of sequence A.

The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — elements of sequence A.

Output

Output on the first line "YES" (without quotes) if it is possible to make sequence A beautiful by performing operations described above, and "NO" (without quotes) otherwise.

If the answer was "YES", output the minimal number of moves needed to make sequence A beautiful.

Examples

Input

2 
1 1

Output

YES 
1

Input

3 
6 2 4

Output

YES 
0

Input

2 
1 3

Output

YES 
1

Note

In the first example you can simply make one move to obtain sequence [0, 2] with $Codeforces Round #410 (Div. 2)(A,字符串,水坑，B,暴力枚举，C,思维题，D,区间贪心)$ .

In the second example the gcd of the sequence is already greater than 1.

题目链接：http://codeforces.com/contest/798/problem/C

分析：

容易发现对于两个数连续做2次之后就都是偶数。2个奇数操作一次就成为偶数。
先对本身所有数求gcd，如果有公共因子则已经成立，不然就按照上述规则，将所有数变为偶数。

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e+7;
 4 int gcd(int a,int b)
 5 {
 6     return b==0?a:gcd(b,a%b);
 7 }
 8 int a[maxn];
 9 int main()
10 {
11     int n,i;
12     cin>>n;
13     for(int i=0;i<n;i++)
14     cin>>a[i];
15     int gccd=a[0];
16     for(i=1;i<n;i++)
17     {
18         gccd=gcd(gccd,a[i]);
19         if(gccd==1)
20             break;
21     }
22     if(i==n)
23     {
24         cout<<"YES"<<endl<<0<<endl;
25         return 0;
26     }
27     else
28     {
29         int c=0,cnt=0;
30         a[n]=0;
31         for(i=0;i<=n;i++)
32         {
33             if(a[i]%2!=0)
34                 cnt++;//求这组数中有多少个奇数
35             else
36             {
37                 c+=(cnt/2);
38                 cnt%=2;
39                 if(cnt!=0)
40                     c+=2;//一个奇数要变2次到偶数
41                 cnt=0;
42             }
43         }
44         cout<<"YES"<<endl<<c<<endl;
45     }
46     return 0;
47 }

Mike has always been thinking about the harshness of social inequality. He's so obsessed with it that sometimes it even affects him while solving problems. At the moment, Mike has two sequences of positive integers A = [a₁, a₂, ..., a_n] and B = [b₁, b₂, ..., b_n] of length n each which he uses to ask people some quite peculiar questions.

To test you on how good are you at spotting inequality in life, he wants you to find an "unfair" subset of the original sequence. To be more precise, he wants you to select k numbers P = [p₁, p₂, ..., p_k] such that 1 ≤ p_i ≤ n for 1 ≤ i ≤ k and elements in P are distinct. Sequence P will represent indices of elements that you'll select from both sequences. He calls such a subset P "unfair" if and only if the following conditions are satisfied: 2·(a_p₁ + ... + a_{p_k}) is greater than the sum of all elements from sequence A, and 2·(b_p₁ + ... + b_{p_k}) is greater than the sum of all elements from the sequence B. Also, k should be smaller or equal to $Codeforces Round #410 (Div. 2)(A,字符串,水坑，B,暴力枚举，C,思维题，D,区间贪心)$ because it will be to easy to find sequence P if he allowed you to select too many elements!

Mike guarantees you that a solution will always exist given the conditions described above, so please help him satisfy his curiosity!

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the number of elements in the sequences.

On the second line there are n space-separated integers a₁, ..., a_n (1 ≤ a_i ≤ 10⁹) — elements of sequence A.

On the third line there are also n space-separated integers b₁, ..., b_n (1 ≤ b_i ≤ 10⁹) — elements of sequence B.

Output

On the first line output an integer k which represents the size of the found subset. k should be less or equal to $Codeforces Round #410 (Div. 2)(A,字符串,水坑，B,暴力枚举，C,思维题，D,区间贪心)$ .

On the next line print k integers p₁, p₂, ..., p_k (1 ≤ p_i ≤ n) — the elements of sequence P. You can print the numbers in any order you want. Elements in sequence P should be distinct.

Example

Input

5 8 7 4 8 3 4 2 5 3 7

Output

3 1 4 5

题目链接：http://codeforces.com/contest/798/problem/D

题意：

给定两个数组大小为n的数组a,b，让你从里面拿k个，使得对于每个数组，拿的元素之和的二倍大于整个数组的元素之和，其中k<=n/2+1

分析：

实在是太套路的一个题目……只描述奇数的情况，偶数随便取一个数之后就变成了奇数的情况。按A排序，取第一个。之后下标2k,2k+1两个数取其中b大的。按这种构造方法得到的即为符合题意的解。

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e+6;
 4 struct node
 5 {
 6     int a,b,j;
 7 }p[maxn];
 8 bool cmp(node c,node d)
 9 {
10     if(c.a!=d.a)
11         return c.a>d.a;
12     else return c.b>d.b;
13 }
14 int main()
15 {
16     int n;
17     while(scanf("%d",&n)!=EOF)
18     {
19         for(int i=1;i<=n;i++)
20         {
21             scanf("%d",&p[i].a);
22             p[i].j=i;
23         }
24         for(int i=1;i<=n;i++)
25             scanf("%d",&p[i].b);;
26         sort(p+1,p+1+n,cmp);
27         printf("%d\n",n/2+1);
28         printf("%d ",p[1].j);
29         for(int i=1;i<=(n-1)/2;i++)
30         {
31             if(p[2*i].b>p[2*i+1].b)
32                 printf("%d ",p[2*i].j);
33             else printf("%d ",p[2*i+1].j);
34         }
35         if(n%2==0)
36             printf("%d ",p[n].j);
37         printf("\n");
38     }
39     return 0;
40 }

Codeforces Round #410 (Div. 2)(A,字符串,水坑，B,暴力枚举，C,思维题，D,区间贪心)

B. Mike and strings

C. Mike and gcd problem

D. Mike and distribution