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;
vector<pair<int,int> >v[15];
int n,m,ans,b[65536],dp[65536][15];
int main(){
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int x,y,len;cin>>x>>y>>len;x--;y--;
		v[x].push_back(make_pair(y,len));v[y].push_back(make_pair(x,len));
		ans+=len; 
	}
	for(int i=0;i<(1<<n);i++)
	{
		for(int j=0;j<n;j++)
			if(!(i&(1<<j))&&!b[i+(1<<j)])
			{
				b[i+(1<<j)]=b[i];
				for(int k=0;k<v[j].size();k++)
					if(i&(1<<v[j][k].first))
						b[i+(1<<j)]+=v[j][k].second;
			}
	}
	memset(dp,-1,sizeof(dp));
	dp[1][0]=0;
	for(int i=0;i<(1<<n);i++)
		for(int j=0;j<n;j++)
			if(dp[i][j]!=-1&&(i&1<<j)){
				for(int k=0;k<v[j].size();k++)
					if(!(i&(1<<v[j][k].first)))
						dp[i|(1<<v[j][k].first)][v[j][k].first]=max(dp[i|(1<<v[j][k].first)][v[j][k].first],dp[i][j]+v[j][k].second);
				for(int k=((1<<n)-1-i)|(1<<j);k;k=(k-1)&(((1<<n)-1-i)|(1<<j)))
					if(k&(1<<j))
						dp[i|k][j]=max(dp[i|k][j],dp[i][j]+b[k]);
			}
	cout<<ans-dp[(1<<n)-1][n-1];
}