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 MAXN = 17, MAXS = 32775;

int n, m, N, g[MAXN][MAXN], val[MAXS], dp[MAXS][MAXN];

inline void update(int &a, int b){if(a < b) a = b;}
int main(){
	int i, j, k, r, rest, u, v, w, sum = 0;
	scanf("%d%d", &n, &m);
	N = 1 << n;
	for(i = 1; i <= m; ++ i){
		scanf("%d%d%d", &u, &v, &w);
		-- u, -- v;
		g[u][v] = g[v][u] = w;
		sum += w;
	} memset(dp, -1, sizeof(dp));
	for(i = 1; i < N; ++ i){
		for(j = 0; j < n; ++ j)
			if(i & (1 << j)) break;
		val[i] = val[i ^ (1 << j)];
		for(k = j + 1; k < n; ++ k)
			if(i & (1 << k))
				val[i] += g[j][k];
		if(j == 0) dp[i][0] = val[i];
	}
	for(i = 1; i < N; ++ i)
		for(j = 0; j < n; ++ j){
			if((v = dp[i][j]) < 0) continue;
			for(k = 0; k < n; ++ k){
				if(i & (1 << k)) continue;
				if(g[j][k] == 0) continue;
				update(dp[i ^ (1 << k)][k], v + g[j][k]);
				for(rest = r = (N - 1) ^ i ^ (1 << k); r; r = (r - 1) & rest)
					update(dp[i ^ r ^ (1 << k)][k], v + val[r ^ (1 << k)] + g[j][k]);
			}
		}
	cout << sum - dp[N - 1][n - 1] << endl;
	return 0;
}