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;

typedef long long LL;
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

int main(){
	int N,M; cin >> N >> M;
	int s = 0;
	vector<vector<int>> c(N,vector<int>(N,0));
	repp(i,0,M){
		int x,y,z; cin >> x >> y >> z;
		--x; --y;
		c[x][y] = c[y][x] = z;
		s += z;
	}
	vector<int> d(1<<N,0);
	repp(k,1,1<<N){
		int i = 0;
		while(!(k&(1<<i))) ++i;
		d[k] = d[k^(1<<i)];
		repp(j,i+1,N) if(k&(1<<j)) d[k] += c[i][j];
	}
	vector<vector<int>> dp(1<<N,vector<int>(N,-123456789));
	dp[1][0] = 0;
	repp(k,0,1<<(N-1)) repp(i,0,N-1) if((k*2+1)&(1<<i)) repp(j,1,N) if(!((k*2+1)&(1<<j))) if(c[i][j] > 0) dp[(k*2+1)^(1<<j)][j] = max(dp[(k*2+1)^(1<<j)][j],dp[k*2+1][i]+c[i][j]);
	repp(k,0,1<<(N-1)){
		for(int m = k ; m > 0 ; m = (m-1)&k){
			dp[2*k+1][N-1] = max(dp[2*k+1][N-1],dp[2*(k^m)+1][N-1]+d[m*2+1]);
			repp(i,0,N-1) if((m^k)&(1<<i)) dp[2*k+1][N-1] = max(dp[2*k+1][N-1],dp[2*(k^m)+1][N-1]+d[m*2+(1<<(i+1))]);
		}
	}
	cout << s-dp[(1<<N)-1][N-1] << endl;
	return 0;
}