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;
const int N=17;

int n,m,G[N][N],g[1<<N],f[1<<N][N],sum=0;

int main() {
	cin>>n>>m;
	for(int i=1,a,b,c;i<=m;i++) {
		cin>>a>>b>>c;--a,--b;sum+=c;
		G[a][b]=G[b][a]=c;
		for(int j=0;j<(1<<n);j++)
			if((j>>a&1)&&(j>>b&1)) g[j]+=c;
	}
	memset(f,-1,sizeof(f));
	f[1][0]=0;
	for(int i=1;i<(1<<n);i++)
		for(int j=0;j<n;j++)
			if(~f[i][j]) {
				for(int k=0;k<n;k++)
					if(!(i>>k&1)&&G[j][k]) f[i|(1<<k)][k]=max(f[i|(1<<k)][k],f[i][j]+G[j][k]);
				for(int all=((1<<n)-1)^i,k=all;k;k=(k-1)&all)
					f[i|k][j]=max(f[i|k][j],f[i][j]+g[k|(1<<j)]);
			}
	cout<<sum-f[(1<<n)-1][n-1];
	return 0;
}