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<algorithm>
#include<cstdlib>
#include<cstring>
#define U ((1<<n)-1)
using namespace std;

const int N = 16;
int a[N+2][N+2];
int cost[(1<<N)+2];
int dp[(1<<N)+2][N+2];
int n,m;

void update(int &x,int y) {x = max(x,y);}

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1; i<=m; i++)
	{
		int x,y,z; scanf("%d%d%d",&x,&y,&z); x--; y--;
		a[x][y] = a[y][x] = z;
	}
	for(int i=0; i<=U; i++)
	{
		cost[i] = 0;
		for(int j=0; j<n; j++)
		{
			if((i&(1<<j))==0) continue;
			for(int k=j+1; k<n; k++)
			{
				if((i&(1<<k))==0) continue;
				cost[i] += a[j][k];
			}
		}
	}
	memset(dp,-1,sizeof(dp));
	dp[1][0] = 0;
	for(int i=1; i<=U; i+=2)
	{
		for(int j=0; j<n; j++)
		{
			if(dp[i][j]==-1) continue;
			if((i&(1<<j))==0) continue;
			for(int k=0; k<n; k++)
			{
				if(a[j][k]==0) continue;
				if(i&(1<<k)) continue;
				update(dp[i|(1<<k)][k],dp[i][j]+a[j][k]);
			}
			int c = (U-i)|(1<<j);
			for(int k=c; k>=0; k=(k==0 ? -1 : (k-1)&c))
			{
				update(dp[i|k][j],dp[i][j]+cost[k]);
			}
		}
	}
	int ans = 0;
	for(int i=(U>>1)+2; i<=U; i+=2) ans = max(ans,dp[i][n-1]);
	printf("%d\n",cost[U]-ans);
	return 0;
}