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:
#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define INF 1000000007
using namespace std;

int n,m,dp[100010][20],val[100010];
int num[20][20];
vector<int> v;

signed main()
{
	ios::sync_with_stdio(false);
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		int a,b,c;
		cin>>a>>b>>c;
		num[a][b]=num[b][a]=c;
	}
	val[0]=0;
	for(int i=1;i<(1<<n);i++){
		v.clear();
		for(int j=1;j<=n;j++) if(i&(1<<(j-1))){
			v.push_back(j); 
		}
		val[i]=val[i^(1<<(v[0]-1))];
		for(int j=1;j<v.size();j++){
			val[i]+=num[v[0]][v[j]];
		}
	}
	for(int i=0;i<(1<<n);i++) for(int j=1;j<=n;j++) dp[i][j]=-INF;
	for(int i=0;i<(1<<n);i++) if(i&1) dp[i][1]=val[i];
	for(int i=1;i<(1<<n);i++){
		for(int j=1;j<=n;j++) if(i&(1<<(j-1))){
			if(dp[i][j]==-INF) continue;
			int xx=((1<<n)-1)^i;
			for(int mask=xx;mask>0;mask=(mask-1)&xx){
				for(int k=1;k<=n;k++) if(mask&(1<<(k-1))&&num[j][k]){
					dp[i^mask][k]=max(dp[i^mask][k],dp[i][j]+num[j][k]+val[mask]);
				}
			}
		}
	}
	cout<<val[(1<<n)-1]-dp[(1<<n)-1][n]<<endl;
	return 0;
}