Codeforces 437C The Child and Toy 简单图论贪心

C. The Child and Toy

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 vi. The child spend vf1 + vf2 + ... + vfk energy for removing part i where f1, f2, ..., fk 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 nintegers: v1, v2, ..., vn (0 ≤ vi ≤ 105). Then followed m lines, each line contains two integers xi and yi, representing a rope from part xi to part yi (1 ≤ xi, yi ≤ n; xi ≠ yi).

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.

一个图n个点m条边，每个点有权值，每次删点的代价是与这个点直接相连的所有点的权值和，求最小权值和。

我们考虑任意一条边：这条边被删，计入答案的必然是两点之一。

那么，我们只要先删权值最大的，就可以使每条边对答案的贡献为两点中权值小的，从而总答案最小。

#include <cstdio>#include <iostream>#include <string.h>#include <string> #include <map>#include <queue>#include <vector>#include <set>#include <algorithm>#include <math.h>#include <cmath>#include <bitset>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))using namespace std;typedef long long ll;typedef long double ld;const int maxn=1005,inf=0x3f3f3f3f;  const ll llinf=0x3f3f3f3f3f3f3f3f;   const ld pi=acos(-1.0L);  int a[maxn];int main() {int n,m,x,y,i;scanf("%d%d",&n,&m);for (i=1;i<=n;i++) {scanf("%d",&a[i]);}int ans=0;for (i=1;i<=m;i++) {scanf("%d%d",&x,&y);if (a[x]<a[y]) ans+=a[x]; else ans+=a[y];}printf("%d\n",ans);return 0;}