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;

int dp[1<<15][15];
int g[15][15];
int dis[1<<15];
int n,m,ans=0x3f3f3f3f,sum;

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n>>m;
    for(int i=0;i<m;i++){
        int a,b,c;
        cin>>a>>b>>c;a--;b--;
        g[a][b]=g[b][a]=c;
        sum+=c;
    }
    for(int s=0;s<1<<n;s++){
        for(int i=0;i<n;i++)if(s>>i&1){
            for(int j=i+1;j<n;j++)if(s>>j&1){
                dis[s]+=g[i][j];
            }
        }
    }
    memset(dp,0x3f,sizeof(dp));
    dp[1][0]=sum;
    for(int s=0;s<1<<n;s++){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++)if(!(s>>j&1)&&g[i][j]){
                dp[s|(1<<j)][j]=min(dp[s|(1<<j)][j],dp[s][i]-g[i][j]);
            }
            int rs=(s^((1<<n)-1))|(1<<i);
            for(int j=rs;j;j=rs&(j-1)){
                dp[s|j][i]=min(dp[s|j][i],dp[s][i]-dis[j]);
            }
        }
    }
    for(int s=0;s<1<<n;s++){
        ans=min(ans,dp[s][n-1]);
    }
    cout<<ans<<endl;

    return 0;
}