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 update(a,b,c) if(dp[a][b]<c)dp[a][b]=c,q.push({a,b})
using namespace std;
struct edge{int v,w;};
int n,m,dp[70001][21],sum,cost[70001];
vector<edge> w[21];
int main()
{
	scanf("%d%d",&n,&m);
	while(m--)
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		x--;y--;
		w[x].push_back({y,z});
		w[y].push_back({x,z});
		sum+=z;
	}
	for(int s=0;s<(1<<n);s++)
		for(int i=0;i<n;i++)
			if(s&(1<<i))
				for(edge j:w[i])
					if(s&(1<<j.v))cost[s]+=j.w;
	for(int s=0;s<(1<<n);s++)cost[s]/=2;
	memset(dp,-1,sizeof dp);
	dp[1][0]=0;
	queue<pair<int,int> > q;
	q.push({1,0});
	while(!q.empty())
	{
		int s=q.front().first,u=q.front().second,ss=((1<<n)-1)^s;q.pop();
		for(edge j:w[u])
			if(~s&(1<<j.v))update(s^(1<<j.v),j.v,j.w+dp[s][u]);
		for(int t=ss;t;t=(t-1)&ss)
			if(~t&(1<<u))update(s^t,u,cost[t^(1<<u)]+dp[s][u]);
	}
	int ans=1<<30;
	for(int i=0;i<(1<<n);i++)ans=min(ans,sum-dp[i][n-1]);
	printf("%d\n",ans);
	return 0;
}