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>
#define N 20
#define N2 65634
using namespace std;

int V, E;
int i, u, v, w;
int G[N][N];
int f[N2][N];
int in[N2], out[N][N2], ans;

void up(int &x, const int y) {x < y ? x = y : 0;}

int main(){
    scanf("%d%d", &V, &E);
    for(i = 0; i < E; ++i){
        scanf("%d%d%d", &u, &v, &w); --u; --v;
        G[u][v] = G[v][u] = w; ans += w;
    }
    for(i = 0; i < 1 << V; ++i)
        for(u = 0; u < V; ++u) if(i >> u & 1)
            for(v = u + 1; v < V; ++v) if(i >> v & 1)
                in[i] += G[u][v];
    for(i = 0; i < 1 << V; ++i)
        for(u = 0; u < V; ++u) if(!(i >> u & 1))
            for(v = 0; v < V; ++v) if(i >> v & 1)
                out[u][i] += G[u][v];
    memset(f, 128, sizeof f);
    f[1][0] = 0;
    for(i = 1; i < 1 << V; ++i)
        for(u = 0; u < V; ++u) if((i >> u & 1) && f[i][u] >= 0){
            for(v = 0; v < V; ++v) if(!(i >> v & 1) && G[u][v])
                up(f[i | 1 << v][v], f[i][u] + G[u][v]);
            v = ~(-1 << V | i);
            for(; v; v = v - 1 & ~i)
                up(f[i | v][u], f[i][u] + out[u][v] + in[v]);
        }
    printf("%d\n", ans -= f[(1 << V) - 1][V - 1]);
    return 0;
}