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;
typedef long long ll;
const ll MOD = 1e9 + 7;
const int INF = 1 << 30;
int mat[15][15];
int cost[1 << 15][15];
int dp[1 << 15][15];
int main() {
	int N, M;
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		a--; b--;
		mat[a][b] = mat[b][a] = c;
	}
	for (int mask = 0; mask < (1 << N); mask++) {
		for (int i = 0; i < N; i++) {
			dp[mask][i] = INF;
			for (int j = 0; j < N; j++) {
				if (mask&(1 << j)) cost[mask][i] += mat[i][j];
			}
		}
	}
	dp[1][0] = 0;
	for (int mask = 0; mask < (1 << N); mask++) {
		for (int i = 0; i < N; i++) {
			if (mask&(1 << i) == 0 || dp[mask][i] == INF) continue;
			for (int j = 0; j < N; j++) {
				if (mask&(1 << j) || mat[i][j] == 0) continue;
				int add = cost[mask][j] - mat[i][j];
				dp[mask | (1 << j)][j] = min(dp[mask | (1 << j)][j], dp[mask][i] + add);
			}
			int nmask = mask ^ (1 << i);
			int comp = ((1 << N) - 1) ^ mask;
			for (int bit = comp; bit > 0; bit = (bit - 1)&comp) {
				int add = 0;
				for (int j = 0; j < N; j++) {
					if (bit&(1 << j)) add += cost[nmask][j];
				}
				dp[mask | bit][i] = min(dp[mask | bit][i], dp[mask][i] + add);
			}
		}
	}
	cout << dp[(1 << N) - 1][N - 1] << endl;
}