cf437C The Child and Toy 贪心

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C. The Child and Toy

time limit per test

1 second

memory limit per test

256 megabytes

On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy the toy.

The toy consists of n parts and m ropes. Each rope links two parts, but every pair of parts is linked by at most one rope. To split the toy, the child must remove all its parts. The child can remove a single part at a time, and each remove consume an energy. Let's define an energy value of part i as v_i. The child spend v_f1 + v_f2 + ... + v_fk energy for removing part i where f₁, f₂, ..., f_k are the parts that are directly connected to thei-th and haven't been removed.

Help the child to find out, what is the minimum total energy he should spend to remove alln parts.

Input

The first line contains two integers n andm (1 ≤ n ≤ 1000;0 ≤ m ≤ 2000). The second line containsn integers: v₁, v₂, ..., v_n (0 ≤ v_i ≤ 10⁵). Then followed m lines, each line contains two integersx_i andy_i, representing a rope from partx_i to party_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i).

Consider all the parts are numbered from 1 to n.

Output

Output the minimum total energy the child should spend to remove all n parts of the toy.

Examples

Input

4 310 20 30 401 41 22 3

Output

40

Input

4 4100 100 100 1001 22 32 43 4

Output

400

Input

7 1040 10 20 10 20 80 401 54 74 55 25 76 41 61 34 31 4

Output

160

Note

One of the optimal sequence of actions in the first sample is:

• First, remove part 3, cost of the action is 20.
• Then, remove part 2, cost of the action is 10.
• Next, remove part 4, cost of the action is 10.
• At last, remove part 1, cost of the action is 0.

So the total energy the child paid is 20 + 10 + 10 + 0 = 40, which is the minimum.

In the second sample, the child will spend 400 no matter in what order he will remove the parts.

翻译：

N个点M条边， 每个点又对应的权值， 断开某一个边和点的连接的花费是这个点的权值， 问断开至任意两点之间不相连的状态的需要的最小权值之和。

分析：

任意两点之间不相连， 等价于断开任意一条边。我们上面说过断开某一条边的花费是这个边和相连的点的权值， 那么既然要求最小， 那么我们对于每一条边， 都选择权值最小的边去断开。

代码：

#include<iostream>  #include<stdlib.h>  #include<stdio.h>  #include<cmath>  #include<algorithm>  #include<string>  #include<string.h>  #include<set>  #include<queue>  #include<stack>  #include<functional>   using namespace std;const int maxn = 20000 + 10;int m, n;int x, y;int v[maxn], c[maxn];int sum;int main(){while (scanf("%d %d", &n, &m )!= EOF) {sum = 0;for (int i = 1; i <= n; i++)scanf("%d", &v[i]);for (int i = 1; i <= m; i++) {scanf("%d %d", &x, &y);sum += min(v[x], v[y]);}printf("%d\n", sum);}//system("pause");return 0;}