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 <cstdio>
#include <algorithm>
using namespace std;
int N,M;
int C[16][16];
int dp[1<<15][16];
int F[1<<15][16];
int wut(int a,int b)
{
    int rez=0;
    for(int i=1;i<=N;i++)if((b>>(i-1))&1)rez+=F[a][i];
    return rez;
}
int main()
{
    scanf("%d %d",&N,&M);
    for(int i=1;i<=M;i++){int x,y,z;scanf("%d %d %d",&x,&y,&z);C[x][y]=C[y][x]=z;}
    for(int conf=0;conf<(1<<N);conf++)
    {
        for(int j=1;j<=N;j++)
        {
            dp[conf][j]=1<<30;
            for(int k=1;k<=N;k++)
                if((conf>>(k-1))&1)F[conf][j]+=C[k][j];
        }
    }
    dp[1][1]=0;
    for(int conf=1;conf<(1<<N);conf++)
    {
        for(int i=1;i<=N;i++)
        {
            if((conf>>(i-1))&1)
            {
                for(int j=1;j<=N;j++)
                    if(C[i][j]&&((conf>>(j-1))&1)==0)
                        dp[conf|(1<<(j-1))][j]=min(dp[conf|(1<<(j-1))][j],dp[conf][i]+F[conf^(1<<(i-1))][j]);
                int u=(1<<N)-1-conf;
                for(int p=u;p;p=(p-1)&u)
                    dp[conf|p][i]=min(dp[conf|p][i],dp[conf][i]+wut(conf^(1<<(i-1)),p));
            }
        }
    }
    printf("%d",dp[(1<<N)-1][N]);
    return 0;
}