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 rep(i, N) for (int i = 0; i < N; i++)
#define pb push_back
 
typedef long long ll;
typedef pair<int, int> i_i;
 
const int MOD = 1e9 + 7;
const ll INF = LLONG_MAX / 2;
 
 
int main() {
	int N, M; cin >> N >> M;
	vector<int> a(M), b(M), c(M);
	rep(i, M) cin >> a[i] >> b[i] >> c[i], a[i]--, b[i]--;
	vector<vector<int> > adj(N, vector<int>(N));
	rep(i, M) adj[a[i]][b[i]] = adj[b[i]][a[i]] = c[i];
	
	vector<int> C(1<<N);
	rep(S, 1<<N) rep(i, M) if ((S>>a[i] & 1) && (S>>b[i] & 1)) C[S] += c[i];
	vector<vector<int> > dp(1<<N, vector<int>(N, -1));
	dp[1][0] = 0;
	rep(S, 1<<N) rep(u, N) if (dp[S][u] != -1) {
		int _S = (1<<N) - 1 - S;
		rep(v, N) if ((_S>>v & 1) && adj[u][v])
			dp[S | 1 << v][v] = max(dp[S | 1 << v][v], dp[S][u] + adj[u][v]);
		for (int T = _S; T; T = (T - 1) & _S)
			dp[S | T][u] = max(dp[S | T][u], dp[S][u] + C[T | 1<<u]);
	}
	int ans = -dp.back()[N - 1];
	rep(i, M) ans += c[i];
	cout << ans << endl;
}