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<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
long long n,m,w[15][15],p[1<<15],dp[1<<15][15],sum,ans;
void init()
{
    for(int i=0;i<1<<n;++i)
        for(int j=0;j<n;++j) if(i>>j&1)
            for(int k=j+1;k<n;++k) if(i>>k&1)
                p[i]+=w[j][k];
}
int main()
{
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=1;i<=m;++i)
    {
        int x,y,t;
        cin>>x>>y>>t;
        --x;--y;
        w[x][y]=t;
        w[y][x]=t;
        sum+=t;
    }
    init();
    memset(dp,0xff,sizeof(dp));
    dp[1][0]=0;
    for(int i=0;i<1<<n;++i)
        for(int j=0;j<n;++j) if(dp[i][j]!=-1)
        {
            int x=(1<<n)-1^i|1<<j;
            for(int y=x;y>=0;--y)
                y&=x,dp[i|y][j]=max(dp[i|y][j],dp[i][j]+p[y]);
            for(int k=0;k<n;++k) if(!(i>>k&1)&&w[j][k]!=0)
                dp[i|1<<k][k]=max(dp[i|1<<k][k],dp[i][j]+w[j][k]);
        }
    for(int i=0;i<1<<n;++i) ans=max(ans,dp[i][n-1]);
    cout<<sum-ans<<endl;
    return 0;
}