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 <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
const int N = 15 + 5, inf = 1e9 + 7 ;
using namespace std ;
int n, m, g[N][N], f[N][(1 << 15) + 5], G[(1 << 15) + 5][N] ; 
inline void upd(int &x, int y) { if (y < x) x = y ; }
int main() {
	scanf("%d%d", &n, &m) ;
	int x, y, z ;
	rep(i, 1, m) {
		scanf("%d%d%d", &x, &y, &z) ;
		g[x][y] = g[y][x] = z ;
	}
	m = (1 << n) - 1 ;
	rep(i, 1, m) rep(j, 1, n) if ((i >> (j - 1)) & 1) rep(k, 1, n) G[i][k] += g[j][k] ;
	rep(i, 1, n) rep(j, 1, m) f[i][j] = inf ; 
	f[1][1] = 0 ;
	rep(j, 1, m) {
	 	rep(i, 1, n) if ((j >> (i - 1)) & 1) rep(x, 1, n) if (!((j >> (x - 1)) & 1) && g[x][i]) 
			upd(f[x][j | (1 << (x - 1))], f[i][j] + G[j][x] - g[x][i]) ;
		int st = m ^ j ;
		for (int x = st ; x; x = (x - 1) & st) {
			int cost = 0 ;
			rep(i, 1, n) if ((x >> (i - 1)) & 1) cost += G[j][i] ;
			rep(i, 1, n) upd(f[i][j ^ x], f[i][j] + cost - G[x][i]) ;
		}
	}
	printf("%d\n", f[n][m]) ;
	return 0 ; 
}