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>
#define rep(i, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
using std::cin; using std::cout;
const int maxn = 15;
int n, m;
int f[maxn][1 << maxn];
int edge[maxn][maxn];
int g[1 << maxn];
inline void up(int & x,int y) {
	if(x < y) {
		x = y;
	}
}
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> m;
	int sum = 0;
	rep(i, 1, m) {
		int a, b, c;
		cin >> a >> b >> c;
		-- a, -- b;
		edge[a][b] = c;
		edge[b][a] = c;
		for(int j = 0;j < 1 << n;++j) if(j >> a & j >> b & 1) {
			g[j] += c;
		}
		sum += c;
	}
	for(int i = 0;i < n;++i) {
		for(int j = 0;j < 1 << n;++j) {
			f[i][j] = -1e9;
		}
	}
	for(int i = 1;i < 1 << n;i += 2)
		f[0][i] = g[i];
	for(int j = 0;j < 1 << n;++j) {
		for(int i = 0;i < n;++i) if(f[i][j] >= 0) {
			for(int l = 0;l < n;++l) if(edge[i][l] && !(j >> l & 1)) {
				for(int s = ~(j | 1 << l) & (1 << n) - 1, k = s;k;k = k - 1 & s) {
					up(f[l][k | j | 1 << l], f[i][j] + edge[i][l] + g[k | 1 << l]);
				}
				up(f[l][j | 1 << l], f[i][j] + edge[i][l]);
			}
		}
	}
	cout << sum - f[n - 1][(1 << n) - 1] << '\n';
}