codeforces444A

DZY Loves Physics

 CodeForces - 444A 

DZY loves Physics, and he enjoys calculating density.

Almost everything has density, even a graph. We define the density of a non-directed graph (nodes and edges of the graph have some values) as follows:

where v is the sum of the values of the nodes, e is the sum of the values of the edges.

Once DZY got a graph G, now he wants to find a connected induced subgraph G' of the graph, such that the density of G' is as large as possible.

An induced subgraph G'(V', E') of a graph G(V, E) is a graph that satisfies:

• ;
• edge  if and only if , and edge ;
• the value of an edge in G' is the same as the value of the corresponding edge in G, so as the value of a node.

Help DZY to find the induced subgraph with maximum density. Note that the induced subgraph you choose must be connected.

Input

The first line contains two space-separated integers n (1 ≤ n ≤ 500), . Integer n represents the number of nodes of the graph G, mrepresents the number of edges.

The second line contains n space-separated integers x_i (1 ≤ x_i ≤ 10⁶), where x_irepresents the value of the i-th node. Consider the graph nodes are numbered from 1to n.

Each of the next m lines contains three space-separated integers a_i, b_i, c_i (1 ≤ a_i < b_i ≤ n; 1 ≤ c_i ≤ 10³), denoting an edge between node a_i and b_i with value c_i. The graph won't contain multiple edges.

Output

Output a real number denoting the answer, with an absolute or relative error of at most 10^- 9.

Examples

Input

1 0
1

Output

0.000000000000000

Input

2 1
1 2
1 2 1

Output

3.000000000000000

Input

5 6
13 56 73 98 17
1 2 56
1 3 29
1 4 42
2 3 95
2 4 88
3 4 63

Output

2.965517241379311

Note

In the first sample, you can only choose an empty subgraph, or the subgraph containing only node 1.

In the second sample, choosing the whole graph is optimal.

sol：超有趣的结论题。

结论：答案肯定是一个最多只有一条边的子图；

证明：在已经有两个点的一张子图中，如果加入一个新点会使得答案更优，那么肯定新点和两个点其中一个或图中一个点比三个点的更优

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0');    return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
const int N=505;
int n,m;
double Quanz[N];
int main()
{
    int i;
    double ansV=0,ansE=0;
    R(n); R(m);
    for(i=1;i<=n;i++) R(Quanz[i]);
    for(i=1;i<=m;i++)
    {
        int x,y;
        double z;
        R(x); R(y); R(z);
        if((ansE==0)||(double)(ansV/ansE)<(double)(Quanz[x]+Quanz[y])/z)
        {
            ansV=Quanz[x]+Quanz[y]; ansE=z;
        }
    }
    if(ansE==0) puts("0.00000000000000000");
    else printf("%.17lf
",ansV/ansE);
    return 0;
}
/*
Input
1 0
1
Output
0.000000000000000

Input
2 1
1 2
1 2 1
Output
3.000000000000000

Input
5 6
13 56 73 98 17
1 2 56
1 3 29
1 4 42
2 3 95
2 4 88
3 4 63
Output
2.965517241379311
*/