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 MAXN 15
#define MAXM 505
#define LL long long
#define P int(1e9+7)
using namespace std;

int N,M;
int G[MAXN][MAXN];

int cost[1<<15];
int dp[MAXN][1<<15];

int main(){
	
	cin>>N>>M;
	int u,v,w;
	int tot = 0;
	for(int i=1;i<=M;i++){
		cin>>u>>v>>w;
		--u; --v;
		G[u][v] = G[v][u] = w;
		tot += w;
	}
	
	for(int s=1;s<(1<<N);s++){
		for(u=0;u<N;u++){
			if(((s>>u)&1)==0) continue; 
			for(int v=u+1;v<N;v++){
				if(((s>>v)&1)==0) continue; 
				cost[s] += G[u][v];
			}
		}
		//cout<<s<<" "<<cost[s]<<endl;
	}
	
	int ans = 0;
	memset(dp, 0xf3, sizeof(dp));

	dp[0][1] = 0;
	int U = (1<<N) - 1;
	
	for(int s=1;s<(1<<N);s++){
			for(u=0;u<N;u++){
			if(((s>>u)&1)==0 || dp[u][s]<0) continue;
			//cout<<u<<" "<<s<<" "<<dp[u][s]<<endl;
			if(u==N-1) ans = max(ans, dp[N-1][s]);
			
			for(v=0;v<N;v++){
				if(((s>>v)&1)==0 && G[u][v]){
					dp[v][s|(1<<v)] = max(dp[v][s|(1<<v)], dp[u][s] + G[u][v]);
				}
			}
			
			int s1 = (U^s)|(1<<u);
			
			for(int s2=s1;s2>0;s2=(s2-1)&s1){
				dp[u][s|s2] = max(dp[u][s|s2], dp[u][s] + cost[s2]);
			}
		}
	}
	cout<<tot - ans<<endl;
	
	return 0;
}