Codeforces Round #250 (Div. 2) D. The Child and Zoo 并查集

D. The Child and Zoo

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Of course our child likes walking in a zoo. The zoo has n areas, that are numbered from 1 to n. The i-th area contains ai animals in it. Also there are m roads in the zoo, and each road connects two distinct areas. Naturally the zoo is connected, so you can reach any area of the zoo from any other area using the roads.

Our child is very smart. Imagine the child want to go from area p to area q. Firstly he considers all the simple routes from p to q. For each route the child writes down the number, that is equal to the minimum number of animals among the route areas. Let's denote the largest of the written numbers as f(p, q). Finally, the child chooses one of the routes for which he writes down the value f(p, q).

After the child has visited the zoo, he thinks about the question: what is the average value of f(p, q) for all pairs p, q (p ≠ q)? Can you answer his question?

Input

The first line contains two integers n and m (2 ≤ n ≤ 105; 0 ≤ m ≤ 105). The second line contains n integers: a1, a2, ..., an(0 ≤ ai ≤ 105). Then follow m lines, each line contains two integers xi and yi (1 ≤ xi, yi ≤ n; xi ≠ yi), denoting the road between areas xiand yi.

All roads are bidirectional, each pair of areas is connected by at most one road.

Output

Output a real number — the value of .

The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4.

Examples

input

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

output

16.666667

input

3 3
10 20 30
1 2
2 3
3 1

output

13.333333

input

7 8
40 20 10 30 20 50 40
1 2
2 3
3 4
4 5
5 6
6 7
1 4
5 7

output

18.571429

Note

Consider the first sample. There are 12 possible situations:

• p = 1, q = 3, f(p, q) = 10.
• p = 2, q = 3, f(p, q) = 20.
• p = 4, q = 3, f(p, q) = 30.
• p = 1, q = 2, f(p, q) = 10.
• p = 2, q = 4, f(p, q) = 20.
• p = 4, q = 1, f(p, q) = 10.

Another 6 cases are symmetrical to the above. The average is .

Consider the second sample. There are 6 possible situations:

• p = 1, q = 2, f(p, q) = 10.
• p = 2, q = 3, f(p, q) = 20.
• p = 1, q = 3, f(p, q) = 10.

Another 3 cases are symmetrical to the above. The average is .

题意：给你一个图，n个点，m条边，sigma f（p，q）/(n*(n-1));q!=p;f（p，q）=点p到点q经过最小的点权值；

思路：将点权值从大到小排序，每次加入一个点，相对应的所加的边的最小值为加入点权值的最小值，并查集处理；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=1e5+,M=4e6+,inf=1e9+;
struct is
{
    int u,v;
    double w;
    bool operator <(const is &b)const
    {
        return w>b.w;
    }
}edge[N];
double v[N];
int father[N],si[N];
int findd(int x)
{
    return x==father[x]?x:father[x]=findd(father[x]);
}
void uni(int u,int v)
{
    int x=findd(u);
    int y=findd(v);
    if(x!=y)
    {
        father[x]=y;
        si[y]+=si[x];
    }
}
int main()
{
    int y,z,i,t;
    ll x;
    scanf("%lld%d",&x,&y);
    for(i=; i<=x; i++)
        father[i]=i,si[i]=;
    for(i=; i<=x; i++)
        scanf("%lf",&v[i]);
    for(i=; i<=y; i++)
    {
        scanf("%d%d",&edge[i].u,&edge[i].v);
        edge[i].w=min(v[edge[i].u],v[edge[i].v]);
    }
    sort(edge+,edge+y+);
    double ans=0.0,minn=10000000.0;
    for(i=; i<=y; i++)
    {
        minn=min(minn,edge[i].w);
        int u=findd(edge[i].u);
        int v=findd(edge[i].v);
        if(u!=v)
        {
            ans+=minn*si[u]*si[v];
            uni(u,v);
        }
    }
    printf("%f\n",ans*/(x*(x-)));
    return ;
}