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 <algorithm>
#include <cstdio>
int a[15][15],f[1<<15][15],g[1<<15];
int main(){
	int n,m,sum=0;
	scanf("%d%d\n",&n,&m);
	for (int i=1;i<=m;i++){
		int x,y,z;
		scanf("%d%d%d\n",&x,&y,&z);
		x--;y--;sum+=z;
		a[x][y]=a[y][x]=z;
	}
	int L=1<<n;
	for (int i=0;i<L;i++){
		for (int j=0;j<n;j++) f[i][j]=-1e9;
		for (int j=0;j<n;j++)
		if (i>>j&1){
			g[i]=g[1<<j^i];
			for (int k=j+1;k<n;k++)
				if (i>>k&1) g[i]+=a[j][k];
			break;
		}
	}
	f[1][0]=0;
	for (int i=1;i<L;i++)
		for (int j=0;j<n;j++)
		if (f[i][j]>=0){
			int x=(L-1)^i;
			for (int k=0;k<n;k++)
				if (!(i>>k&1) && a[j][k])
					f[i|1<<k][k]=std::max(f[i|1<<k][k],f[i][j]+a[j][k]);
			for (int k=x;k;k=(k-1)&x)
				f[i|k][j]=std::max(f[i|k][j],f[i][j]+g[k|1<<j]);
		}
	printf("%d\n",sum-f[L-1][n-1]);
}