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 int long long 
using namespace std;
int n,m,x,y,z,a[20][20],kk,f[20][40000],p[40000],sum;
inline int min(int x,int y){return x<y?x:y;}
inline int max(int x,int y){return x>y?x:y;}
signed main(){
	ios::sync_with_stdio(false);
	cin>>n>>m;for (int i=1;i<=m;i++){
		cin>>x>>y>>z;if (x>y)swap(x,y);a[x][y]+=z;sum+=z;
	}kk=1<<n;for (int l=1;l<kk;l++)
		for (int i=1;i<=n;i++)if (l&(1<<(i-1)))
			for (int j=1;j<=n;j++)if (l&(1<<(j-1)))p[l]+=a[i][j];
	memset(f,200,sizeof(f));
	for (int i=1;i<kk;i++)if (i&1)f[1][i]=p[i];
	for (int i=1;i<kk;i++)
		for (int x1=1;x1<=n;x1++)if (i&(1<<(x1-1))){
			int P=(kk-1)^i;
			for (int x2=1;x2<=n;x2++)if (a[min(x1,x2)][max(x1,x2)])
				for (int j=P;j;j=(j-1)&P)if (j&(1<<(x2-1)))
					f[x2][i|j]=max(f[x2][i|j],f[x1][i]+a[min(x1,x2)][max(x1,x2)]+p[j]);
		}cout<<sum-f[n][kk-1]<<endl;
}