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>
#define maxn 20
using namespace std;
int n,m,dis[maxn][maxn],sum;
int dp[1<<maxn][maxn],s[1<<maxn];
int main() {
	scanf("%d %d",&n,&m);
	memset(dis,0,sizeof(dis));
	sum=0;
	for (int i=1;i<=m;i++) {
		int a,b,c;
		scanf("%d %d %d",&a,&b,&c);
		a--; b--;
		dis[a][b]=c;
		dis[b][a]=c;
		sum+=c;
	}
	for (int i=0;i<(1<<n);i++) {
		for (int j=0;j<n;j++)
			for (int k=j+1;k<n;k++)
				if ((i>>j)%2==1&&(i>>k)%2==1)
					s[i]+=dis[j][k];
	}
	memset(dp,-1,sizeof(dp));
	dp[1][0]=0;
	for (int i=0;i<(1<<n);i++) {
		for (int j=0;j<n;j++) {
			if (dp[i][j]==-1)
				continue;
			for (int t=0;t<n;t++) {
				if ((i>>t)%2==1||dis[j][t]==0)
					continue;
				dp[i|(1<<t)][t]=max(dp[i|(1<<t)][t],dp[i][j]+dis[j][t]);
			}
			int x=(1<<n)-1-i;
			for (int y=x;y>=1;y=(y-1)&x)
				dp[i|y][j]=max(dp[i|y][j],dp[i][j]+s[y|(1<<j)]);
		}
	}
	printf("%d\n",sum-dp[(1<<n)-1][n-1]);
	return 0;
}