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>
using namespace std;
const int MaxN=15;
int n,m,ans,e[MaxN][MaxN],c[1<<MaxN],dp[1<<MaxN][MaxN];
int main() {
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;++i) {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        u--,v--;
        e[u][v]=e[v][u]=w;
        ans+=w;
    }
    for (int i=0;i<(1<<n);++i) for (int j=0;j<n;++j) for (int k=j+1;k<n;++k) if (i&(1<<j) && i&(1<<k)) c[i]+=e[j][k];
    memset(dp,0xff,sizeof dp);dp[1][0]=0;
    for (int S=0;S<(1<<n);++S) for (int i=0;i<n;++i) if (dp[S][i]!=-1) {
        int SS=((1<<n)-1)^S;
        for (int j=0;j<n;++j) if (SS&(1<<j) && e[i][j]) 
            dp[S|(1<<j)][j]=max(dp[S|(1<<j)][j],dp[S][i]+e[i][j]);
        for (int j=SS;j;j=(j-1)&SS) 
            dp[S|j][i]=max(dp[S|j][i],dp[S][i]+c[j|(1<<i)]);
    }
    ans-=dp[(1<<n)-1][n-1];
    printf("%d\n",ans);
    return 0;
}