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;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()
#define dump(x) cout << #x << " = " << (x) << endl

const int MN = 15;
const int MB = 1 << MN;
const ll inf = -1e9;

int e[MN][MN];
int ct[MB];
int dp[MN][MB];

int main() {
	int N, M; cin >> N >> M;
	int sum = 0;
	rep(i, M) {
		int a, b, c;
		cin >> a >> b >> c;
		--a; --b;
		e[a][b] = e[b][a] = c;
		sum += c;
	}

	rep(i, N) rep(j, 1 << N) {
		dp[i][j] = inf;
	} 

	rep(j, 1 << N) {
		rep(u, N) rep(v, u) {
			if (((j >> u) & 1) && ((j >> v) & 1)) {
				ct[j] += e[u][v];
			}
		}
	}

	rep(j, 1 << N) {
		if (j & 1) {
			dp[0][j] = ct[j];
		}
	}

	rep(j, 1 << N) {
		for (int k = j; k > 0; k = (k - 1) & j) {
			rep(i, N) if (((j ^ k) >> i) & 1) {
				rep(p, N) if (((k >> p) & 1) && dp[p][k] != inf && e[p][i] > 0) {
					dp[i][j] = max(dp[i][j], dp[p][k] + ct[j ^ k] + e[p][i]);
				}
			}
		}
	}

	int ans = sum - dp[N-1][(1 << N) - 1];
	cout << ans << endl;
	return 0;
}