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 INF = 1e9;

const int MAXN = 15;
int N;
int W[MAXN][MAXN];

int V[1<<MAXN];
int dp[1<<MAXN][MAXN];

int main() {
	ios::sync_with_stdio(0), cin.tie(0);
	int M;
	cin >> N >> M;
	int tot = 0;
	for (int i = 0; i < M; i++) {
		int a, b, c; cin >> a >> b >> c; a--, b--;
		W[a][b] = W[b][a] = c;
		tot += c;
	}

	for (int m = 0; m < (1<<N); m++) {
		for (int i = 0; i < N; i++) {
			for (int j = i+1; j < N; j++) {
				if ((m & (1 << i)) && (m & (1 << j))) {
					V[m] += W[i][j];
				}
			}
		}
	}

	for (int m = 0; m < (1<<N); m++) {
		for (int i = 0; i < N; i++) {
			dp[m][i] = -INF;
		}
	}
	dp[1][0] = 0;
	for (int m = 1; m < (1<<N); m++) {
		for (int i = 0; i < N; i++) {
			if (dp[m][i] == -INF) continue;

			for (int j = 0; j < N; j++) {
				if (m & (1 << j)) continue;
				if (W[i][j] == 0) continue;
				dp[m | (1 << j)][j] = max(dp[m | (1 << j)][j], dp[m][i] + W[i][j]);
			}

			int c = (1 << N) - 1 - m;
			for (int v = c; v; v = (v-1) & c) {
				dp[m|v][i] = max(dp[m|v][i], dp[m][i] + V[v | (1 << i)]);
			}
		}
	}
	cout << tot - dp[(1<<N)-1][N-1] << '\n';

	return 0;
}