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 pb push_back
#define mp make_pair
#define LL long long
#define INF 0x3f3f3f3f
#define Inf 1000000000000000000LL
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
int n,m;
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n>>m;
	vector<vector<int> >a(n,vector<int>(n,0));
	LL total=0;
	for(int i=0;i<m;i++){
		int x,y,z;
		cin>>x>>y>>z;
		x--,y--;
		a[x][y]=a[y][x]=z;
		total+=z;
	}
	const int inf=(int)1e9;
	vector<int>inside(1<<n);
	for(int t=0;t<(1<<n);t++)
		for(int i=0;i<n;i++)if(t&(1<<i))
			for(int j=i+1;j<n;j++)if(t&(1<<j))
				inside[t]+=a[i][j];
	vector<vector<int> >dp(1<<n,vector<int>(n,-1));
	for(int t=0;t<(1<<n);t++)if(t&1)dp[t][0]=inside[t];
	for(int t=0;t<(1<<n);t++)
		for(int ii=0;ii<n;ii++){
			if(dp[t][ii]==-1)continue;
			int u=(1<<n)-1-t;
			while(u>0){
				for(int j=0;j<n;j++)
					if(u&(1<<j))if(a[ii][j]>0)
							dp[t|u][j]=max(dp[t|u][j],dp[t][ii]+inside[u]+a[ii][j]);
				u=(u-1)&((1<<n)-1-t);
			}
		}
	cout<<total-dp[(1<<n)-1][n-1]<<endl;
	return 0;
}