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:
//created by missever

#include<bits/stdc++.h>
#define MAX 1000000007
using namespace std;
typedef long long LL;
typedef pair<int,int> P;

int f[1 << 15][15],g[1 << 15];
int p[15][15];

void add(int &x,int y) {
    x = max(x,y);
}

int main() {
    int n,m,i,j,k,u,v,w,s = 0;
    scanf("%d%d",&n,&m);
    for(i = 0; i < m; i++) {
        scanf("%d%d%d",&u,&v,&w);
        u--;
        v--;
        p[u][v] = p[v][u] = w;
        s += w;
    }
    memset(f,-1,sizeof(f));
    for(i = 0; i < (1 << n); i++) {
        for(j = 0; j < n; j++) {
            if(!((i >> j) & 1)) continue;
            for(k = j + 1; k < n; k++) {
                if((i >> k) & 1) g[i] += p[j][k];
            }
        }
        if(i & 1) f[i][0] = g[i];
    }
    for(i = 0; i < (1 << n); i++) {
        for(j = 0; j < n; j++) {
            if(f[i][j] != -1) {
                for(k = 0; k < n; k++) {
                    if(((i >> k) & 1) || !p[j][k]) continue;
                    add(f[i | (1 << k)][k],f[i][j] + p[j][k]);
                    u = ((1 << n) - 1) ^ i ^ (1 << k);
                    for(v = u; v; v = (v - 1) & u) add(f[i | (1 << k) | v][k],f[i][j] + p[j][k] + g[v | (1 << k)]);
                }
            }
        }
    }
    printf("%d\n",s - f[(1 << n) - 1][n - 1]);
    return 0;
}