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>
using namespace std;

#define go(i,a,b) for(int i=a;i<=b;++i)
#define com(i,a,b) for(int i=a;i>=b;--i)
#define mem(a,b) memset(a,b,sizeof(a))
#define fo(i,a) for(int i=0;i<a;++i)
#define il inline

const int inf=0x3f3f3f3f,N=4e4;

int n,m,block[N],dp[N][20],g[20][20];

il void read(int &x){
	x=0;char c=getchar(),f=1;
	while(!isdigit(c)){ if(c=='-') f=-1; c=getchar(); }
	while(isdigit(c)){ x=x*10+c-'0'; c=getchar(); }
	x*=f;
}
il void gmax(int &x,const int &y){
	x=x>y?x:y;
}

int main(){
	//freopen("input.txt","r",stdin);
	read(n),read(m);
	int x,y,w;
	go(i,1,m){
		read(x),read(y),read(w);
		--x,--y;
		g[x][y]=g[y][x]=w;
	}
	int all=(1<<n)-1;
	go(i,1,all)
		fo(j,n){
			if(i>>j&1){
				block[i]=block[i^1<<j];
				fo(k,n) if(i>>k&1) block[i]+=g[j][k];
				break;
			}
		}
	mem(dp,-1);
	dp[1][0]=0;
	go(i,1,all)
		fo(j,n)
			if(i>>j&1&&dp[i][j]!=-1){
				fo(k,n)
					if(!(i>>k&1)&&g[j][k])
						gmax(dp[i|1<<k][k],dp[i][j]+g[j][k]);
				int tmp=(all^i)|1<<j;
				for(int k=tmp;k;k=(k-1)&tmp){
					if(k>>j&1)
						gmax(dp[i|k][j],dp[i][j]+block[k]);
				}
			}
	cout<<block[all]-dp[all][n-1];
	return 0;
}