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=1e5+10;
int n,m,go[20][20],dp[N][20],bit[N],q1[20],q2[20];
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=0;i<n;i++) bit[1<<i]=i;
	for (int i=1;i<=m;i++){
		int u,v,l;
		scanf("%d%d%d",&u,&v,&l);u--;v--;
		go[u][v]=go[v][u]=l;
	}
	int to=1<<(n-1);
	for (int S=0;S<to;S++){
		for (int i=0;i<n;i++) dp[S][i]=1e9;
		for (int T=(S-1)&S;T<S;T=(T-1)&S){
			int s1=0,s2=0,v=0;
			for (int now=S^T;now;now-=now&-now) q2[++s2]=bit[now&-now];//可能的起点
			for (int now=T|to;now;now-=now&-now) q1[++s1]=bit[now&-now];//可能的终点
			for (int i=1;i<=s1;i++) for (int j=1;j<=s2;j++) v+=go[q1[i]][q2[j]];
			for (int i=1;i<=s1;i++) for (int j=1;j<=s2;j++) if (go[q1[i]][q2[j]])
				dp[S][q2[j]]=min(dp[S][q2[j]],dp[T][q1[i]]+v-go[q1[i]][q2[j]]);
		}
		dp[S][n-1]=0;
	}
	printf("%d\n",dp[to-1][0]);
	return 0;
}