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 <cstdio>
using namespace std;
const int INF=0x3f3f3f3f;
int a[15][15],val[1<<15];
int dp[1<<15][15];
void tense(int &x,int y){
	if (x<y)
		x=y;
}
int main(){
	int n,m,sum=0;
	scanf("%d%d",&n,&m);
	while(m--){
		int u,v,c;
		scanf("%d%d%d",&u,&v,&c);
		u--,v--;
		for(int i=0;i<(1<<n);i++)
			if ((i>>u&1) && (i>>v&1))
				val[i]+=c;
		a[u][v]=a[v][u]=c;
		sum+=c;
	}
	for(int i=0;i<(1<<n);i++)
		for(int j=0;j<n;j++)
			dp[i][j]=-INF;
	for(int i=0;i<(1<<n);i++)
		if (i&1){
			dp[i][0]=val[i];
			//printf("val[%d]=%d\n",i,val[i]);
		}
	for(int i=0;i<(1<<n);i++)
		for(int j=0;j<n;j++) 
			if (dp[i][j]>=0){
				//puts("WTF");
				int w=~i&((1<<n)-1);
				for(int k=w;k;k=(k-1)&w){
					int to=k|i;
					for(int o=0;o<n;o++)
						if ((k>>o&1) && a[j][o])
							tense(dp[to][o],dp[i][j]+val[k]+a[j][o]);
				}
			}
	printf("%d",sum-dp[(1<<n)-1][n-1]);
}