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 N 19
#define M 35009
using namespace std;
 
int n,m,a[N][N],f[M],g[M][N],dp[M][N];
int main(){
	scanf("%d%d",&n,&m);
	int i,j,k,x,y,z;
	for (i=1; i<=m; i++){
		scanf("%d%d%d",&x,&y,&z);
		a[x][y]=a[y][x]=z;
		f[1<<x-1|1<<y-1]+=z;
	}
	int tot=1<<n;
	for (i=1; i<=n; i++)
		for (j=1; j<tot; j++) if (j>>i-1&1) f[j]+=f[j^1<<i-1];
	memset(dp,195,sizeof(dp)); memset(g,195,sizeof(g));
	int ans=-1000000000;
	for (i=1; i<tot; i+=2){
		if (!(i>>n-1&1)) dp[i][1]=f[i];
		for (j=1; j<=n; j++) if (i>>j-1&1){
			for (k=i-1&i; k; k=k-1&i) if (k&1)
				dp[i][j]=max(dp[i][j],g[k][j]+f[i^k]);
			if (i>>n-1&1){ if (j==n) ans=max(ans,dp[i][j]); continue; }
			for (k=1; k<=n; k++) if (!(i>>k-1&1) && a[j][k]>0)
				g[i][k]=max(g[i][k],dp[i][j]+a[j][k]);
		}
	}
	printf("%d\n",f[tot-1]-ans);
	return 0;
}