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;

int n, m;

vector <int> g[15], c[15];
void add_edge(int u, int v, int w) {
	g[u].push_back(v);
	c[u].push_back(w);
}

int dp[1 << 15][15];
int prep[1 << 15][15];

void upd(int &x, int y) {
	if (y < x) x = y;
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 0; i < m; ++ i) {
		int u, v, w;
		scanf("%d %d %d", &u, &v, &w);
		-- u; -- v;
		add_edge(u, v, w);
		add_edge(v, u, w);
	}
	memset(dp, 0x3f, sizeof(dp));
	int full = (1 << n) - 1;
	for (int msk = 1; msk < (1 << n); ++ msk) {
		for (int i = 0; i < n; ++ i) if (!(msk >> i & 1)) {
			for (int e = 0; e < g[i].size(); ++ e) {
				if (msk >> g[i][e] & 1) prep[msk][i] += c[i][e];
			}
		}
	}
	dp[1][0] = 0;
	for (int msk = 1; msk < (1 << n); ++ msk) {
		for (int u = 0; u < n; ++ u) if (msk >> u & 1) {
			for (int e = 0; e < g[u].size(); ++ e) {
				int v = g[u][e];
				if (!(msk >> v & 1))
					upd(dp[msk | 1 << v][v], dp[msk][u] + prep[msk ^ 1 << u][v]);
			}
			for (int s = full ^ msk; s; s = (s - 1) & (full ^ msk)) {
				int cur = 0;
				for (int i = 0; i < n; ++ i) if (s >> i & 1)
					cur += prep[msk ^ 1 << u][i];
				upd(dp[msk | s][u], dp[msk][u] + cur);
			}
		}
	}
	printf("%d\n", dp[full][n - 1]);
	return 0;
}