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 <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int inf = 0x7fffffff;
int f[1<<15][15];
int w[1<<15][15];
int map[15][15];
int n, m;
void up(int &x, int y) { if(x < y) x = y; }
int main() {
	int i, j, k;
	scanf("%d%d", &n, &m);
	int sum = 0;
	for(i = 1; i <= m; i++){
		int x, y, d;
		scanf("%d%d%d", &x, &y, &d); x--; y--;
		map[x][y] = map[y][x] = d;
		sum += d;
	}
	for(i = 1; i < 1<<n; i++){
		int res = 0;
		for(j = 0; j < n; j++){
			for(k = j+1; k < n; k++) if(i&(1<<j) && i&(1<<k)) res += map[j][k];
		}
		for(j = 0; j < n; j++){
			if(i&(1<<j)) continue;
			w[i][j] = res;
			for(k = 0; k < n; k++){
				if(i&(1<<k)) w[i][j] += map[j][k];
			}
		}
	}
	for(j = 0; j < 1<<n; j++) for(i = 0; i < n; i++) f[j][i] = -inf;
	f[1][0] = 0;
	for(i = 1; i < 1<<n; i += 2){
		for(j = 0; j < n; j++){
			if(!(i&(1<<j))) continue;
			for(k = 0; k < n; k++){
				if(i&(1<<k) && j != k && map[j][k] != 0) up(f[i][j], f[i^(1<<j)][k]+map[j][k]);
			}
			for(k = (i-1)&i; k; k = (k-1)&i) if(k&(1<<j)) up(f[i][j], f[k][j]+w[i^k][j]);
		}
	}
	printf("%d\n", sum-f[(1<<n)-1][n-1]);
	return 0;
}