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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;

LL e[25][25] = {0};
LL W[1 << 16] = {0};
LL dp[1 << 16][25] = {0};
LL n,m,tot = 0;

int main(){
	cin >> n >> m;
	for(LL i = 1;i <= m;i ++){
		LL u,v,w;
		cin >> u >> v >> w;
		e[u][v] = w;
		e[v][u] = w;
		tot += w;
	}
	for(LL S = 0;S < (1LL << n);S ++){
		for(LL i = 1;i <= n;i ++){
			for(LL j = i + 1;j <= n;j ++){
				if(((S >> (i - 1)) & 1) && ((S >> (j - 1)) & 1)) W[S] += e[i][j];
			}
		}
	}
	memset(dp,-0x3f,sizeof(dp));
	dp[1][1] = 0;
	for(LL S = 0;S < (1LL << n);S ++){
		for(LL u = 1;u <= n;u ++){
			if(!((S >> (u - 1)) & 1)) continue;
			for(LL v = 1;v <= n;v ++){
				if((S >> (v - 1)) & 1) continue;
				if(!e[u][v]) continue;
				dp[S | (1 << (v - 1))][v] = max(dp[S | (1 << (v - 1))][v],dp[S][u] + e[u][v]);
			}
			LL U = ((1LL << n) - 1) ^ S;
			for(LL T = U;T;T = (U & (T - 1))){
				dp[S | T][u] = max(dp[S | T][u],dp[S][u] + W[T | (1 << (u - 1))]);
			}
		}
	}
	cout << tot - dp[(1LL << n) - 1][n] << endl;
	return 0;
}