B. Segment Occurrences

You are given two strings t, both consisting only of lowercase Latin letters.

The substring sl,sl+1,…,sr without changing the order.

Each of the occurrences of string a).

You are asked s[li..ri].

Input

The first line contains three integer numbers tand the number of queries, respectively.

The second line is a string |s|=n), consisting only of lowercase Latin letters.

The third line is a string |t|=m), consisting only of lowercase Latin letters.

Each of the next i-th query.

Output

Print s[li..ri].

Examples

input

Copy

10 3 4
codeforces
for
1 3
3 10
5 6
5 7

output

Copy

input

Copy

15 2 3
abacabadabacaba
ba
1 15
3 4
2 14

output

Copy

4
0
3

input

Copy

3 5 2
aaa
baaab
1 3
1 1

output

Copy

0
0

Note

In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.

给两个字符串s，t，之后给出s的一个区间，问这个子串中存在多少个子串与t相同

前缀和相减,但是又带有一定技巧.

 1 #include <bits/stdc++.h>
 2 #define ll long long int
 3 #define inf 0x3f3f3f3f
 4 #define N 1000005
 5 using namespace std;
 6 int a[1005];
 7 int n,m,k;
 8 int main(){
 9     cin>>n>>m>>k;
10     string s,ss;
11     cin>>s>>ss;
12     for(int i=0;i<=n-m;i++){
13         bool flag = true;
14         for(int j=0;j<m;j++){
15             if(s[i+j]!=ss[j]){
16                 flag = false;
17                 break;
18             }
19         }
20         if(flag){
21             a[i+1]++;
22         }
23     }
24     for(int i=1;i<=n;i++){
25         a[i] += a[i-1];
26     }
27     for(int i=0;i<k;i++){
28         int x,y;
29         cin>>x>>y;
30         y = y-m+1;
31         if(x>y){
32             cout<<"0"<<endl;
33         }else{
34             cout<<a[y]-a[x-1]<<endl;
35         }
36     }
37     return 0;
38 }