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;

const int N = 15, inf = 1e9;

int w[N][N], n, m, prep[N][1<<N], dp[N][1<<N]; 

int cost(int s, int t){
	int res = 0, mask = 0;
	for(int i = 0; i < n; ++i) mask += (1 << i) * ((s >> i) & 1);
	for(int i = 0; i < n; ++i) res += prep[i][mask] * ((t >> i) & 1);
	return res;
}

signed main(){
	memset(dp, 0x3f, sizeof(dp));
	
	cin >> n >> m;
	for(int i = 0; i < m; ++i){
		int u, v, c;
		cin >> u >> v >> c, --u, --v;
		w[u][v] = w[v][u] = c;
	}
	
	for(int i = 0; i < n; ++i){
		for(int mask = 0; mask < (1<<n); ++mask){
			for(int j = 0; j < n; ++j){
				if(mask & (1<<j)) prep[i][mask] += w[i][j];
			}
		}
	}
	
	dp[0][1] = 0;
	for(int mask = 1; mask < (1<<n); ++mask){
		for(int i = 0; i < n; ++i){
			if(!(mask & (1 << i))){
				for(int j = 0; j < n; ++j){
					if(mask & (1 << j) && w[j][i]){
						dp[i][mask | (1 << i)] = min(dp[i][mask | (1 << i)], dp[j][mask] + cost(mask ^ (1 << j), (1 << i)));
					}
				}
			}
		}
		int comp = (1<<n) - mask;
		for(int sub = comp; sub > 0; sub = (sub - 1) & comp){
			for(int i = 0; i < n; ++i){
				if(mask & (1 << i)){
					dp[i][mask | sub] = min(dp[i][mask | sub], dp[i][mask] + cost(mask ^ (1 << i), sub));
				}
			}
		}
	}

	cout << dp[n-1][(1<<n) - 1];
}