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>
#include<cstring>
#include<algorithm>
using namespace std;

int n,m,map[16][16],x,y,z,mmh[1<<15][15],o[1<<15][15];
inline int work(int a,int b){
	int mmh=0;
	for (int i=0;i<n;i++)
	if ((a>>i)&1) mmh+=o[b][i];
	return mmh;
}
int main(){
	scanf("%d%d",&n,&m);
	memset(map,0,sizeof(map));
	memset(mmh,63,sizeof(mmh));
	while(m--){
		scanf("%d%d%d",&x,&y,&z);x--;y--;
		map[x][y]=map[y][x]=z;
	}
	for (int i=0;i<(1<<n);i++)
	for (int j=0;j<n;j++){
		o[i][j]=0;
		for (int k=0;k<n;k++)
		if ((i>>k)&1) o[i][j]+=map[j][k];
	}
	mmh[1][0]=0;
	for (int i=0;i<(1<<n);i++)
	for (int j=0;j<n;j++)
	if ((i>>j)&1){
		for (int k=0;k<n;k++)
		if (map[j][k])
		if (!((i>>k)&1)) mmh[i|(1<<k)][k]=min(mmh[i|(1<<k)][k],mmh[i][j]+work(i,1<<k)-map[j][k]);
		
		int u=(1<<n)-1-i;
		for (int p=u;p;p=(p-1)&u) mmh[i|p][j]=min(mmh[i|p][j],mmh[i][j]+work(i^(1<<j),p));
	}
	printf("%d\n",mmh[(1<<n)-1][n-1]);
}