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:
#define _GLIBCXX_IOSTREAM
#include<bits/stdc++.h>
#define inf 1000000007
using namespace std;
int f[1<<15],dp[1<<15][15],tot,c[15][15],n,m;
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;++i)
		for(int j=0;j<n;++j)
			c[i][j]=(i==j?0:-inf);
	for(int i=1,u,v;i<=m;++i)
		scanf("%d%d",&u,&v),scanf("%d",&c[--u][--v]),tot+=(c[v][u]=c[u][v]);
	for(int i=0;i<(1<<n);++i)
		for(int j=0;j<n;++j)if(i>>j&1)
			for(int k=j;k<n;++k)if(i>>k&1)if(c[j][k]!=-inf)
				f[i]=max(-inf,f[i]+c[j][k]);
	for(int i=0;i<(1<<n);++i)
		for(int j=0;j<n;++j)
			dp[i][j]=-inf;
	dp[1][0]=0;
	for(int i=1;i<(1<<n);++i)
		for(int j=0;j<n;++j)if(dp[i][j]!=-inf){
			int nk=(((1<<n)-1)^i)|(1<<j);
			for(int k=nk;k;k=(k-1)&nk)if(k>>j&1)
				dp[i|k][j]=max(dp[i|k][j],dp[i][j]+f[k]);
			for(int k=0;k<n;++k)if(~i>>k&1)
				dp[i|(1<<k)][k]=max(dp[i|(1<<k)][k],dp[i][j]+c[j][k]);//,printf("{%d->%d}",j,k);
//			printf("dp[%d][%d]=%d\n",i,j,dp[i][j]);
		}
	int ans=0;
	for(int i=1;i<(1<<n);++i)ans=max(ans,dp[i][n-1]);
	printf("%d",tot-ans);
}