CodeForces 437C The Child and Toy

The Child and Toy

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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 the i-th and haven't been removed.

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

Input

The first line contains two integers n and m (1 ≤ n ≤ 1000; 0 ≤ m ≤ 2000). The second line contains n integers: v₁, v₂, ..., v_n (0 ≤ v_i ≤ 10⁵). Then followed m lines, each line contains two integers x_i and y_i, representing a rope from part x_i to part y_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.

Sample Input

Input

4 3
10 20 30 40
1 4
1 2
2 3

Output

40

Input

4 4
100 100 100 100
1 2
2 3
2 4
3 4

Output

400

Input

7 10
40 10 20 10 20 80 40
1 5
4 7
4 5
5 2
5 7
6 4
1 6
1 3
4 3
1 4

Output

160

Hint

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.

思路：....................复杂问题的背后说不定都有一种简单的解决方法（看懂题 什么都好了............）

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    int n,m;
    int a[1005];
    while(cin>>n>>m)
    {
        int i;
        for(i=1;i<=n;i++)
            cin>>a[i];
        int x,y;
        int ans=0;
        while(m--)
        {
            cin>>x>>y;
            ans+=min(a[x],a[y]);
        }
        cout<<ans<<endl;
    }
}