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,i,j,k,l,mp[25][25],all[1<<16],dp[1<<16][25],x,y,z,sum,ans;
int main(){
	cin>>n>>m;
	for (i=1;i<=m;i++){
		cin>>x>>y>>z;
		x--;y--;
		sum+=z;
		mp[x][y]=mp[y][x]=z;
	}
	for (i=0;i<(1<<n);i++){
		for (j=0;j<n;j++)if ((i>>j)&1){
			for (k=j+1;k<n;k++)if ((i>>k)&1){
				all[i]+=mp[j][k];
			}
		}
	}
	memset(dp,-1,sizeof(dp));
	dp[1][0]=0;
	for (i=1;i<(1<<n);i++){
		for (j=0;j<n;j++) if ((i>>j)&1)if (dp[i][j]!=-1){
			for (k=((((1<<n)-1)^i)|(1<<j));k;k=((k-1)&((((1<<n)-1)^i)|(1<<j)))){
				dp[i|k][j]=max(dp[i|k][j],dp[i][j]+all[k]);
			}
			for (k=0;k<n;k++) if (!((i>>k)&1)) if (mp[j][k]){
				dp[i|(1<<k)][k]=max(dp[i|(1<<k)][k],dp[i][j]+mp[j][k]);
			}
		}
		/*if (i>32766){
			system("pause");
		}*/
		if ((i>>(n-1))&1) ans=max(ans,dp[i][n-1]);
		/*if (ans>1234567){
			system("pause");
		}*/
	}
	cout<<sum-ans<<endl;
	return 0;
}