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 <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N=17;
const int M=(1<<15)+2;
int dp[M][N],E[N][N],sum[M],all,n,m;

template <typename _Tp> inline void IN(_Tp&x) {
    char ch;bool flag=0;x=0;
    while(ch=getchar(),!isdigit(ch)) if(ch=='-') flag=1;
    while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
    if(flag) x=-x;
}

inline void addedge(int u=0,int v=0,int w=0)
{IN(u),IN(v),IN(w),--u,--v,all+=w,E[u][v]=E[v][u]=w;}
int main() {
    IN(n),IN(m);
    for(int i=1;i<=m;++i) addedge();
    int lim=1<<n;
    for(int S=1;S<lim;++S)
        for(int i=0;i<n;++i) if(S&(1<<i))
            for(int j=i+1;j<n;++j) if(S&(1<<j))
                sum[S]+=E[i][j];
    for(int S=0;S<lim;++S) for(int t=0;t<n;++t) dp[S][t]=-1e9;
    dp[1][0]=0;
    for(int S=1;S<lim;++S) for(int t=0;t<n;++t) if(S&(1<<t)) {
        for(int u=0;u<n;++u) if(!(S&(1<<u))&&E[t][u])
            dp[S|(1<<u)][u]=max(dp[S|(1<<u)][u],dp[S][t]+E[t][u]);
        int P=(lim-1)^S;
        for(int T=P;T;T=(T-1)&P)
            dp[S|T][t]=max(dp[S|T][t],dp[S][t]+sum[T|(1<<t)]);
    }
    printf("%d\n",all-dp[lim-1][n-1]);
    return 0;
}