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;
const int maxn=100010,mod=998244353;
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define ROF(i,a,b) for(int i=(a);i>=(b);i--)
#define MEM(x,v) memset(x,v,sizeof(x))
inline int read(){
	int x=0,f=0;char ch=getchar();
	while(ch<'0' || ch>'9') f|=ch=='-',ch=getchar();
	while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
	return f?-x:x;
}
int n,m,lim,wsum,e[15][15],f[1<<15][15],tot[1<<15];
inline void chkmax(int &x,int y){if(x<y) x=y;}
int main(){
	n=read();m=read();lim=1<<n;
	FOR(i,1,m){
		int u=read()-1,v=read()-1;
		wsum+=e[u][v]=e[v][u]=read();
	}
	FOR(i,0,lim-1) FOR(j,0,n-1) if((i>>j)&1) FOR(k,j+1,n-1) if((i>>k)&1) tot[i]+=e[j][k];
	MEM(f,~0x3f);
	f[1][0]=0;
	FOR(i,1,lim-1) FOR(j,0,n-1) if((i>>j)&1){
		FOR(k,0,n-1) if(!((i>>k)&1) && e[j][k]) chkmax(f[i|(1<<k)][k],f[i][j]+e[j][k]);
		int re=(lim-1)^i;
		for(int k=re;k;k=((k-1)&re)) chkmax(f[i|k][j],f[i][j]+tot[k|(1<<j)]);
	}
	printf("%d\n",wsum-f[lim-1][n-1]);
}