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;
typedef long long ll;
int n,m;
int c[16][16];
int dp[1<<15][15];
int cost[1<<15];
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--;
        c[u][v]=c[v][u]=w;
    }
    for(int s=0;s<(1<<n);s++){
        for(int i=0;i<n;i++){
            if((s>>i&1)==0)continue;
            for(int j=i+1;j<n;j++){
                if((s>>j&1)==0)continue;
                cost[s]+=c[i][j];
            }
        }
    }
    for(int s=0;s<(1<<n);s++){
        for(int u=0;u<n;u++)dp[s][u]=-1;
    }
    for(int s=0;s<(1<<n);s++)if(s&1)dp[s][0]=cost[s];
    for(int s=0;s<(1<<n);s++){
        for(int u=0;u<n;u++){
            if(dp[s][u]==-1)continue;
            int mask=(1<<n)-1-s;
            for(int ns=mask;ns;ns=(ns-1)&mask){
                for(int v=0;v<n;v++){
                    if((ns>>v&1)&&c[u][v])dp[s|ns][v]=max(dp[s|ns][v],dp[s][u]+c[u][v]+cost[ns]);
                }
            }
        }
    }
    printf("%d\n",cost[(1<<n)-1]-dp[(1<<n)-1][n-1]);
    return 0;
}