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 in(x, S) (S & (1 << (x - 1)))
using namespace std;

const int MAXN = 15 + 10, MAXSTATUS = (1 << 15) + 10;

int n, m, e[MAXN][MAXN], g[MAXSTATUS], f[MAXN][MAXSTATUS], num;

int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1, u, v, w; i <= m; ++i) {
        scanf("%d %d %d", &u, &v, &w), num += w;
        e[u][v] = e[v][u] = w;
    } 
    int MAXS = 1 << n;
    for (int S = 0; S < MAXS; ++S) {
        for (int i = 1; i <= n; ++i) {
            for (int j = i + 1; j <= n; ++j) {
                if (in(i, S) && in(j, S) && e[i][j]) g[S] += e[i][j];
            }
        }
    }
    memset(f, -1, sizeof f);
    f[1][1] = 0;
    for (int S = 1; S < MAXS; ++S) {
        for (int i = 1; i <= n; ++i) if (in(i, S) && f[i][S] != -1) {
            for (int j = 1; j <= n; ++j) {
                if (!in(j, S) && e[i][j]) {
                    f[j][S | (1 << (j - 1))] = max(f[j][S | (1 << (j - 1))], f[i][S] + e[i][j]);
                }
            }
            for (int tmp = (MAXS - 1) ^ S, T = tmp; T; T = (T - 1) & tmp) {
                f[i][S | T] = max(f[i][S | T], f[i][S] + g[T | (1 << (i - 1))]);
            }
        }
    }
    printf("%d\n", num - f[n][MAXS - 1]);
    return 0;
}