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 For(i,l,r) for(int i = (l),i##end = (r);i <= i##end;i++)
#define Fordown(i,r,l) for(int i = (r),i##end = (l);i >= i##end;i--)
using namespace std;
typedef long long ll;
const int INF = 1e9;
const int N = 15;
int dp[1 << N][N + 1],Sum[1 << N],G[N + 1][N + 1];
inline int read(){
	int x = 0,flag = 1;
	char ch = getchar();
	while(!isdigit(ch) && ch != '-')ch = getchar();
	if(ch == '-')flag = -1,ch = getchar();
	while(isdigit(ch))x = (x << 3) + (x << 1) + ch - '0',ch = getchar();
	return x * flag;
}
int main(){
	int n = read(),m = read(),tot = 0;
	For(i,1,m){
		int x = read(),y = read(),z = read();
		G[x][y] = G[y][x] = z,tot += z;
	}
	For(i,1,(1 << n) - 1){
		For(j,1,n)if(i & (1 << (j - 1)))
			For(k,j + 1,n)if(i & (1 << (k - 1)))
				Sum[i] += G[j][k];
	}
	memset(dp,-63,sizeof(dp));
	dp[1][1] = 0;
	For(i,1,(1 << n) - 1)For(j,1,n){
		int res = dp[i][j];
		if(res < 0)continue;
		For(k,1,n)if(!(i & (1 << (k - 1))) && G[j][k])
			dp[i | (1 << (k - 1))][k] = max(dp[i | (1 << (k - 1))][k],res + G[j][k]);
		int S0 = (1 << n) - 1 - i;
		for(int S = S0;S;S = (S - 1) & S0)
			dp[i | S][j] = max(dp[i | S][j],res + Sum[S | (1 << (j - 1))]);
	}
	printf("%d\n",tot - dp[(1 << n) - 1][n]);
	return 0;
}