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 <iostream>
#define llint long long

using namespace std;

llint n, m;
llint G[25][25];
llint sum[1<<15];
llint dp[1<<15][25];

int main(void)
{
	cin >> n >> m;
	llint u, v, w;
	for(int i = 0; i < m; i++){
		cin >> u >> v >> w;
		u--, v--;
		G[u][v] = G[v][u] = w;
	}
	
	int N = 1<<n;
	for(int i = 0; i < N; i++){
		for(int j = 0; j < n; j++){
			for(int k = 0; k < n; k++){
				if((i&(1<<j)) && (i&(1<<k))) sum[i] += G[j][k];
			}
		}
		sum[i] /= 2;
	}
	
	for(int i = 0; i < N; i++){
		for(int j = 0; j < n; j++) dp[i][j] = -1e18;
	}
	dp[1][0] = 0;
	
	for(int i = 0; i < N; i++){
		for(int j = 0; j < n; j++){
			for(int k = 0; k < n; k++){
				if(i&(1<<k)) continue;
				if(G[j][k] > 0) dp[i|(1<<k)][k] = max(dp[i|(1<<k)][k], dp[i][j]+G[j][k]);
			}
			int S = (N-1)-i+(1<<j);
			for(int k = S; k > 0; k = (k-1)&S){
				if((k&(1<<j)) == 0) continue;
				dp[i|k][j] = max(dp[i|k][j], dp[i][j]+sum[k]);
			}
		}
	}
	cout << sum[N-1] - dp[N-1][n-1] << endl;
	
	return 0;
}