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;

#define debug(...) fprintf(stderr, __VA_ARGS__)

const long long inf = 1e18;

int n, m, adj[15 + 5][15 + 5];
long long dp[(1 << 15) + 5][15 + 5];

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; ++i) {
		int x, y, w;
		scanf("%d%d%d", &x, &y, &w), --x, --y;
		adj[x][y] = adj[y][x] = w;
	}
//	for (int i = 0; i < n; ++i) {
//		for (int j = 0; j < n; ++j)
//			debug("%d ", adj[i][j]);
//		debug("\n");
//	}
	for (int i = 0; i < (1 << n); ++i)
		for (int j = 0; j < n; ++j)
			dp[i][j] = inf;
	for (int i = 1; i < (1 << n); i += 2)
		dp[i][0] = 0;
	for (int i = 1; i < (1 << n); i += 2)
		for (int j = i; j; (--j) &= i) {
			if (j & 1) continue;
			int t = i ^ j;
			long long tot = 0;
			for (int k = 0; k < n; ++k) {
				if (~t & 1 << k) continue;
				for (int p = 0; p < n; ++p)
					if (j & 1 << p) tot += adj[k][p];
			}
			for (int l = 0; l < n; ++l) {
				if (~t & 1 << l) continue;
				for (int k = 0; k < n; ++k)
					if (j & 1 << k && adj[l][k])
						dp[i][k] = min(dp[i][k], dp[t][l] + tot - adj[l][k]);
			}
		}
	printf("%lld\n", dp[(1 << n) - 1][n - 1]);
}