make the fence great again(dp 二维)

You have a fence consisting of

Unfortunately, it is possible that now your fence is not great. But you can change it! You can increase the length of the

Calculate the minimum number of rubles you have to spend to make the fence great again!

You have to answer

Input

The first line contains one integer

The first line of each query contains one integers

The following

It is guaranteed that sum of all

It is guaranteed that answer to each query will not exceed

Output

For each query print one integer — the minimum number of rubles you have to spend to make the fence great.

Example

input

Copy

3
3
2 4
2 1
3 5
3
2 3
2 10
2 6
4
1 7
3 3
2 6
1000000000 2

output

Copy

2
9
0

Note

In the first query you have to increase the length of second board by

In the second query you have to increase the length of first board by

In the third query the fence is great initially, so you don't need to spend rubles.

这题。。。比赛的时候居然没有做出来，嘤嘤嘤

 1 /*
 2     本题大意：给定一串数字，现在你要把这串数字变为相邻数字不相等
 3     的一串数字，变化方法为：对某个位置的数+1(可加无限多次)，且有一定的花费，每个
 4     位置花费不同，问最小花费。
 5     本题思路：
 6         本题可以说是想到就很好写的线性dp了，因为不知道某个数字
 7         到底要加几次，也不知道某个数字加了之后会不会影响其它数字
 8         ，我们不可能依次枚举某个数字加的次数，但是我们很清晰的知道
 9         每个数字最多只能加两次，为什么呢？因为加入a[i] 和 a[i + 1]
10         相等，加入这次我们改变a[i]，那么a[i] += 1;加了之后如果形成
11         了a[i - 1] == a[i]，那么我们还可以选择其中较小的花费加1，也
12         就是有可能选择a[i]又加一，此时a[i] 肯定和两端的数字都不一样
13         所以说每个数字必定最多加两次就可以使得他和左右两边的数字都不一样。
14         也就是是说我们可以用记忆化搜索搞定这个题目。
15         我们用dp[i][j]表示第i个数字加了j次使得第i个数字之前的所有数字
16         不矛盾的最小花费，那么很明显。
17         我们用j表示第i-1个数加的次数，k表示第i个数加的次数
18         因此我们判断
19         if val[i] + k != val[i - 1] + j:
20             dp[i][k] = min(dp[i][k], dp[i - 1][j] + k * cost[i]) 
21 */
22 #include <cstdio>
23 #include <cstring>
24 using namespace std;
25 
26 typedef long long ll;
27 const int maxn = 300000 + 5;
28 ll inf =  1e18 + 5;
29 int n;
30 ll a[maxn], b[maxn];
31 ll dp[maxn][3];
32 
33 ll min(ll a, ll b) {
34     return a > b ? b : a;
35 }
36 
37 int main() {
38     int q;
39     scanf("%d", &q);
40     while(q --) {
41         scanf("%d", &n);
42         for(int i = 0; i <= n; i ++) {
43             for(int j = 0; j < 3; j ++) dp[i][j] = inf;
44         }
45         for(int i = 1; i <= n; i ++) {
46             scanf("%lld %lld", a + i, b + i);
47         }
48         dp[0][0] = 0ll;
49         for(int i = 1; i <= n; i ++) {
50             for(int j = 0; j < 3; j ++) {
51                 for(int k = 0; k < 3; k ++) {
52                     if(a[i] + j != a[i - 1] + k) 
53                         dp[i][j] = min(dp[i][j], dp[i - 1][k] + j * b[i]);
54                 }
55             }
56         }
57         printf("%lld\n", min(min(dp[n][0], dp[n][1]), dp[n][2]));
58     }
59     return 0;  
60 }