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 Rint register int
using namespace std;
const int N = 15;
int n, m, lim, wsum, edge[N][N], sum[1 << N], dp[N][1 << N];
inline int cost(int S, int T){return sum[S | T] - sum[S] - sum[T];}
inline bool chkmax(int &a, int b){return (a < b) ? (a = b,1) : 0;}
int main(){
	scanf("%d%d", &n, &m); lim = 1 << n;
	for(Rint i = 1;i <= m;i ++){
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c); -- a; -- b;
		edge[a][b] = edge[b][a] = c; wsum += c;
	}
	for(Rint S = 1;S < lim;S ++)
		for(Rint i = 0;i < n;i ++) if((S >> i) & 1)
			for(Rint j = i + 1;j < n;j ++) if((S >> j) & 1)
				sum[S] += edge[i][j];
	memset(dp, 0x80, sizeof dp);
	dp[0][1] = 0;
	for(Rint S = 1;S < lim;S += 2)
		for(Rint t = 0;t < n;t ++) if((S >> t) & 1){
			for(Rint u = 0;u < n;u ++) if(!((S >> u) & 1) && edge[t][u])
				chkmax(dp[u][S | (1 << u)], dp[t][S] + edge[t][u]);
			int SS = (lim - 1) ^ S;
			for(Rint T = SS;T;T = (T - 1) & SS)
				chkmax(dp[t][S | T], dp[t][S] + sum[T | (1 << t)]);
		}
	printf("%d\n", wsum - dp[n - 1][lim - 1]);
}