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>
using namespace std;
int n,m,g[15][15],dp[1<<15][15],cost[1<<15],sum;
int main() {
	memset(dp,-1,sizeof(dp));
	memset(g,-1,sizeof(g));
	scanf("%d%d",&n,&m);
	while(m--) {
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		g[a-1][b-1]=g[b-1][a-1]=c;
		sum+=c;
	}
	for(int i=1;i<(1<<n);i++)
		for(int j=0;j<n;j++)
			if((1<<j)&i)
				for(int k=j+1;k<n;k++)
					if((1<<k)&i&&g[j][k]>-1)
						cost[i]+=g[j][k];
	for(int i=1;i<1<<n;i+=2) dp[i][0]=cost[i];
	for(int i=1;i<(1<<n);i+=2)
		for(int j=0;j<n;j++)
			if(dp[i][j]>-1)
				for(int k=(1<<n)-1-i;k;k=(k-1)&(1<<n)-1-i)
					for(int l=0;l<n;l++)
						if(1<<l&k&&g[j][l]>-1)
							dp[i|k][l]=max(dp[i|k][l],dp[i][j]+g[j][l]+cost[k]);
	printf("%d\n",sum-dp[(1<<n)-1][n-1]);
	return 0;
}