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 <algorithm>

#define Rep(i, n) for (int i = 1; i <= n; i ++)
#define Rep0(i, n) for (int i = 0; i <= n; i ++)
#define RepG(i, x) for (int i = head[x]; i; i = edge[i].next)

using namespace std;

const int N = 21;
const int N0 = 33000;

int g[N][N], t0[N0], t1[N0][N], n, m;

int f[N][N0][2];

int cut(int s1, int s2){ return t0[s1 | s2] - t0[s1] - t0[s2];}
#define g(i) (1 << (i - 1))
int main()
{
	
	scanf("%d%d", &n, &m);
	Rep(i, m) {
		int x, y, z;
		scanf("%d%d%d", &x, &y, &z);
		g[x][y] += z, g[y][x] += z;
	}
	
	Rep0(i, (1 << n) - 1) {
		Rep(j, n) if (i & (1 << (j - 1))){
			for (int k = j; k <= n; k ++)
				if (i & (1 << (k - 1))) t0[i] += g[j][k];
			Rep(k, n) t1[i][k] += g[k][j];
		}
	}
	
	Rep(i, n) Rep0(s, (1 << n) - 1) f[i][s][0] = f[i][s][1] = 1e9;
	
	f[1][1][0] = 0;
	
	Rep0(s, (1 << n) - 1) Rep(i, n) if (s & g(i)){
		int tmp = s ^ g(i);
		Rep(j, n) if ((tmp & g(j)) && g[i][j])
			f[i][s][0] = min(f[i][s][0], f[j][tmp][1] - g[i][j] + t1[tmp][i]);
		for (int s0 = tmp; ; s0 = tmp & (s0 - 1)) {
			f[i][s][1] = min(f[i][s][1], f[i][s ^ s0][0] + cut(s0, tmp ^ s0));
			if (!s0) break;
		}
	}
	
	printf("%d\n", f[n][(1 << n) - 1][1]);
	
	return 0;
}
/*
5 6
1 5 100
1 4 100
1 2 100
2 3 100
3 4 100
4 5 100

*/