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=15;
int n,m,x,y,z,g[1<<N],a[N][N],f[1<<N][N],ans;
int main(){
	scanf("%d%d",&n,&m);
	int sum=0;
	while (m--){
		scanf("%d%d%d",&x,&y,&z);
		sum+=z;
		x--;y--;
		a[x][y]=a[y][x]=z;
	}
	for (int s=1;s<1<<n;s++)
		for (int i=0;i<n;i++)
			if (s&(1<<i)){
				g[s]=g[s^(1<<i)];
				for (int j=0;j<n;j++)
					if ((s&(1<<j)))g[s]+=a[i][j];
				break;
			}
	memset(f,172,sizeof f);
	f[1][0]=0;
	for (int s=1;s<1<<n;s++)
		for (int j=0;j<n;j++)
			if (f[s][j]>=0){
				for (int i=0;i<n;i++)
					if (!(s&(1<<i))&&a[j][i])
						f[s|(1<<i)][i]=max(f[s|(1<<i)][i],f[s][j]+a[j][i]);
				int sy=((1<<n)-1)^s;
				for (int i=sy;i;i=(i-1)&sy)f[s|i][j]=max(f[s|i][j],f[s][j]+g[i|(1<<j)]);
			}
	int ans=0;
	for (int i=1;i<1<<n;i++)ans=max(ans,f[i][n-1]);
	printf("%d",sum-ans);
}