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<cstdlib>
#include<cstring>
#include<algorithm>
#define MAXN 15
#define INF 0x3F3F3F3F
int v[MAXN][MAXN], n, m, x, y, z;
int c[1 << MAXN][MAXN], f[1 << MAXN][MAXN];
int main(){
	scanf("%d%d", &n, &m);
	for(int i = 0;i < m;i++){
		scanf("%d%d%d", &x, &y, &z);
		x--;y--;
		v[x][y] = v[y][x] = z;
	}
	for(int d = 0;d < (1 << n);d++){
		for(int i = 0;i < n;i++){
			f[d][i] = INF;
			for(int j = 0;j < n;j++)if((d >> j) & 1)c[d][i] += v[i][j];
		}
	}
	f[1][0] = 0;
	for(int d = 0;d < (1 << n);d++){
		for(int i = 0;i < n;i++){
			if(((d >> i) & 1) && f[d][i] != INF){
				for(int j = 0;j < n;j++){
					if(!((d >> j) & 1) && v[i][j]){
						f[d | (1 << j)][j] = std::min(
							f[d | (1 << j)][j],
							f[d][i] + c[d][j] - v[i][j]
						);
					}
				}
				register int next = d ^ (1 << i), st = ((1 << n) - 1) ^ d, sum;
				for(int j = st;j > 0;j = (j - 1) & st){
					sum = 0;
					for(int l = 0;l < n;l++)if((j >> l) & 1)sum += c[next][l];
					f[d | j][i] = std::min(f[d | j][i], f[d][i] + sum);
				}
			}
		}
	}
	printf("%d\n", f[(1 << n) - 1][n - 1]);
	return 0;
}