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 INF 0x3f3f3f3f
const int MOD=1e9+7;
const double eps=1e-8;
typedef long long ll;
//orz yht
using namespace std;
int n,m,edge[15][15],sco[32770];
int dp[15][32770];
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out"，"w",stdout);
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int a,b,c;
		cin>>a>>b>>c;
		a--;
		b--;
		edge[a][b]=edge[b][a]=c;
	}
    for(int i=0;i<(1<<n);i++){
        sco[i]=0;
        for(int j=0;j<n;j++){
            for(int k=j+1;k<n;k++){
                if(i&(1<<j) && i&(1<<k))
                sco[i]+=edge[j][k];
            }
        }
    }
    memset(dp,-INF,sizeof dp);
    for(int i=1;i<(1<<n);i+=2)
    dp[0][i]=sco[i];
    for(int i=0;i<(1<<n);i++){
        for(int j=0;j<n;j++){
            if(dp[j][i]<0)
			continue;
            int msk=(1<<n)-1-i;
            for(int sub=msk;sub>0;sub=(sub-1)&msk){
                for(int k=0;k<n;k++){
                    if(edge[j][k] && (sub&(1<<k)))
                    dp[k][i+sub]=max(dp[k][i+sub],dp[j][i]+sco[sub]+edge[j][k]);
                }
            }
        }
    }
    ll cnt=0;
    for(int i=0;i<n;i++)
    for(int j=i+1;j<n;j++)
    cnt+=edge[i][j];
    cout<<cnt-dp[n-1][(1<<n)-1];
	return 0;
}