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 rep(i,a,b) for (int i=(a); i<=(b); i++)
#define subset(i,S) for (int i=(S); i; i=(i-1)&(S))
using namespace std;

const int maxn = 16;
int f[1<<maxn][maxn], cost[1<<maxn], w[maxn][maxn];
int n, m, a, b, c, S, total;

void init() {
	rep (i, 1, (1<<n)-1) 
		rep (x, 0, n-1) if (i>>x&1)
			rep (y, x+1, n-1) if (i>>y&1)
				cost[i] += w[x][y];
}

int main() {
	scanf("%d%d", &n, &m);
	rep (i, 1, m) {
		scanf("%d%d%d", &a, &b, &c); a--; b--;
		w[a][b] = w[b][a] = c; total += c;
	}
	init();
	memset(f, -1, sizeof f);
	f[1][0] = 0;
	rep (i, 1, (1<<n)-1)
		rep (j, 0, n-1) if ((i>>j&1) && ~f[i][j]) {
			S = (((1<<n)-1)^i)|(1<<j);
			subset (s, S) f[i|s][j] = max(f[i|s][j], f[i][j] + cost[s]);
			rep (k, 0, n-1) if (!(i>>k&1) && w[j][k]) f[i|(1<<k)][k] = max(f[i|(1<<k)][k], f[i][j] + w[j][k]);
		}
	printf("%d\n", total - f[(1<<n)-1][n-1]);
	return 0;
}