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>
int f[17777][17],n,m;
int se[33333],e[16][16],tr[16],sum=0,ans=0;
void maxs(int&a,int b){if(a<b)a=b;}
int main(){
	memset(e,-0x3f,sizeof(e));
	scanf("%d%d",&n,&m);
	for(int i=0,a,b,c;i<m;++i){
		scanf("%d%d%d",&a,&b,&c);
		--a,--b;sum+=c;
		e[a][b]=e[b][a]=c;
		int s=1<<a|1<<b;
		for(int j=0;j<(1<<n);++j)if((j&s)==s)se[j]+=c;
	}
	--n;
	int mx=1<<n;
	memset(f,-0x3f,sizeof(f[0])*mx);
	for(int i=1;i<mx;++i)if(i&1)f[i][0]=se[i];
	for(int i=1;i<mx;++i)if(i&1){
		memset(tr,-0x3f,sizeof(tr));
		for(int j=0;j<n;++j)if(i>>j&1){
			for(int k=0;k<n;++k)if(~i>>k&1){
				maxs(tr[k],f[i][j]+e[j][k]);
			}
		}
		int rv=mx-1-i;
		for(int j=rv;j;j=j-1&rv){
			int v=se[j];
			for(int k=0;k<n;++k)if(j>>k&1)maxs(f[i|j][k],tr[k]+v);
		}
		for(int j=0;j<n;++j)maxs(ans,f[i][j]+e[n][j]+se[mx*2-1-i]);
	}
	printf("%d\n",sum-ans);
	return 0;
}