Problem: Mole decided to live in an abandoned mine. The structure of the mine is represented by a simple connected undirected graph which consists of N vertices numbered 1 through N and M edges. The i-th edge connects Vertices a_i and b_i, and it costs c_i yen (the currency of Japan) to remove it.

Mole would like to remove some of the edges so that there is exactly one path from Vertex 1 to Vertex N that does not visit the same vertex more than once. Find the minimum budget needed to achieve this.

Constraints

* 2 \leq N \leq 15
* N-1 \leq M \leq N(N-1)/2
* 1 \leq a_i, b_i \leq N
* 1 \leq c_i \leq 10^{6}
* There are neither multiple edges nor self-loops in the given graph.
* The given graph is connected.

Input

Input is given from Standard Input in the following format:


N M
a_1 b_1 c_1
:
a_M b_M c_M


Output

Print the answer.

Examples

Input

4 6
1 2 100
3 1 100
2 4 100
4 3 100
1 4 100
3 2 100


Output

200


Input

2 1
1 2 1


Output

0


Input

15 22
8 13 33418
14 15 55849
7 10 15207
4 6 64328
6 9 86902
15 7 46978
8 14 53526
1 2 8720
14 12 37748
8 3 61543
6 5 32425
4 11 20932
3 12 55123
8 2 45333
9 12 77796
3 9 71922
12 15 70793
2 4 25485
11 6 1436
2 7 81563
7 11 97843
3 1 40491


Output

133677

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long
#define For(i,a,b) for (int i=a;i<=b;i++)
#define Rep(i,a,b) for (int i=b;i>=a;i--)
using namespace std;
int n,m,mp[16][16],x,y,z,mmh[1<<15][15],o[1<<15][15];
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
inline int work(int a,int b)
{
    int mmh=0;
    for (int i=0;i<n;i++)
    if ((a>>i)&1) mmh+=o[b][i];
    return mmh;
}
int main()
{
    n=read();m=read();
    memset(mp,0,sizeof(mp));
    memset(mmh,63,sizeof(mmh));
    while(m--)
	{
        x=read();y=read();z=read();
		x--;y--;
        mp[x][y]=mp[y][x]=z;
    }
    For(i,0,(1<<n)-1)
	    For(j,0,n-1)
		{
	        o[i][j]=0;
	        For(k,0,n-1)if ((i>>k)&1) o[i][j]+=mp[j][k];
	    }
    mmh[1][0]=0;
    For(i,0,(1<<n)-1)
	    For(j,0,n-1)
		    if ((i>>j)&1){
		        For(k,0,n-1)
		       		if (mp[j][k])
		        		if (!((i>>k)&1)) 
							mmh[i|(1<<k)][k]=min(mmh[i|(1<<k)][k],mmh[i][j]+work(i,1<<k)-mp[j][k]);
		        int u=(1<<n)-1-i;
		        for (int p=u;p;p=(p-1)&u) mmh[i|p][j]=min(mmh[i|p][j],mmh[i][j]+work(i^(1<<j),p));
		    }
    printf("%d\n",mmh[(1<<n)-1][n-1]);
}