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>
#define N 20
using namespace std;

int n, m, map[N][N], x, y, z, w[32770], bin[N], f[32770][N], ans, s, ss;

int main(){
	scanf("%d%d", &n, &m);
	memset(map, 0, sizeof(map));
	for(int i=1; i<=m; i++){scanf("%d%d%d", &x, &y, &z); map[x][y]=map[y][x]=z;}
	bin[0]=1; for(int i=1; i<=n; i++)bin[i]=bin[i-1]<<1;
	for(int i=0; i<=bin[n]-1; i++){
		w[i]=0;
		for(int j=1; j<=n; j++)if(i&bin[j-1])
			for(int k=j+1; k<=n; k++)if(i&bin[k-1])w[i]+=map[j][k];
	}
	memset(f, -1, sizeof(f));
	for(int i=1; i<=bin[n]-1; i++)if(i&bin[0])
		for(int j=1; j<=n; j++)if(i&bin[j-1]){
			if(j==1){f[i][j]=w[i]; continue;}
			s=i-bin[j-1];
			for(int k=1; k<=n; k++)if(j!=k&&i&bin[k-1]&&map[k][j]){
				x=k>=2?bin[0]+bin[k-1]:bin[0];
				s-=x;
				for(int s1=s; s1>=0; s1=(s1-1)&s){
					ss=s1+x; if(f[ss][k]>=0)f[i][j]=max(f[i][j], f[ss][k]+w[i-ss]+map[k][j]);
					if(!s1)break;
				}
				s+=x;
			}
		}
	ans=-f[bin[n]-1][n];
	for(int i=1; i<=n; i++)
		for(int j=i+1; j<=n; j++)ans+=map[i][j];
	printf("%d", ans);
	return 0;
}