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 e[15][15];
int n;
int dp1[1<<15];
int dp2[1<<15][15];
int tmp[15];
void upd(int&x,int y){
	if(x==-1 || x<y)x=y;
}
int main(){
	int m, tot = 0;
	scanf("%d%d",&n,&m);
	while(m--){
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		--x;--y;
		e[x][y] = e[y][x] = z;
		tot += z;
	}
	for(int mask=1; mask < (1<<n); mask ++){
		int t;
		for(t=0; (mask&(1<<t)) == 0; t++);
		dp1[mask] = dp1[mask-(1<<t)];
		for(int i=t+1; i<n; i++)if(mask&(1<<i))
			dp1[mask] += e[t][i];
	}
	memset(dp2,-1,sizeof(dp2));
	for(int mask=1; mask < (1<<n); mask++)if((mask&1)!=0&&(mask&(1<<(n-1)))==0)
		dp2[mask][0] = dp1[mask];

	for(int mask=1; mask < (1<<n); mask++){
		memset(tmp,-1,sizeof(tmp));
		for(int i=0; i<n; i++)if((mask&(1<<i))!=0 && dp2[mask][i] != -1)
		for(int j=0; j<n; j++)if((mask&(1<<j)) == 0 && e[i][j] != 0)
			upd(tmp[j], dp2[mask][i] + e[i][j]);
		int comp = (1<<n) - 1 - mask, b = 0;
		do{
			for(int i=0; i<n; i++)if((b & (1<<i))!=0 && tmp[i]!=-1){
				upd(dp2[mask + b][i], tmp[i] + dp1[b]);
			}
		}while(b=(b-comp)&comp);
	}
	printf("%d\n",tot - dp2[(1<<n)-1][n-1]);
	return 0;
}