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>
#define ll long long
#define N 20
#define oo 1e9
#define For(i,x,y) for(int i=x;i<=y;i++)
#define Rep(i,x,y) for(int i=x;i<y;i++)
#define mp make_pair
#define pa pair<ll,int>
using namespace std;
inline int read(){
	int x=0,f=0;char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar()) if(ch=='-') f=1;
	for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-48;
	return f?-x:x;
}
int n,m,G[N][N],f[1<<N][N],s[1<<N],sum;
int main(){
	n=read();m=read();
	For(i,1,m){int x=read()-1,y=read()-1;G[x][y]=G[y][x]=read();sum+=G[x][y];}
	Rep(i,0,1<<n) Rep(j,0,n) if((i>>j)&1) Rep(k,j+1,n) if((i>>k)&1) s[i]+=G[j][k];
	Rep(i,0,1<<n) Rep(j,0,n) f[i][j]=-oo;
	Rep(i,0,1<<n) if(i&1) f[i][0]=s[i];
	Rep(i,0,1<<n) Rep(j,0,n) if((i>>j)&1) for(int k=i;k;k=(k-1)&i) if(k&1) Rep(l,0,n) if(((k>>l)&1)&&((~k>>j)&1)&&G[l][j]) f[i][j]=max(f[i][j],f[k][l]+G[l][j]+s[i^k]);
	printf("%d\n",sum-f[(1<<n)-1][n-1]);
	return 0;
}