//ループverです。移動後はvisitedが増えるので, visited昇順にループを回せば良いです。
#include <iostream>
#include <algorithm>
using namespace std;

int INF = 1000000007;
int n, m;
bool canMove[15][15] = {false};
int ecost[15][15];
int dp[1 << 15][15];	//dp[visited][pos] = これまでの最小コスト（visitedの下位i(>=0)ビット目が1⇔頂点iを訪問済み, pos…現在地.）

int main() {
	int i, j, k;
	
	cin >> n >> m;
	for (i = 0; i < m; i++) {
		int s, t, d;
		cin >> s >> t >> d;
		ecost[s][t] = d;
		canMove[s][t] = true;
	}
	
	for (i = 0; i < (1 << n); i++) for (j = 0; j < n; j++) dp[i][j] = INF;
	dp[1][0] = 0;
	for (i = 1; i < (1 << n) - 1; i++) {
		for (j = 0; j < n; j++) {
			if (dp[i][j] >= INF) continue;	//この行は無くてもよい
			for (k = 0; k < n; k++) {
				if (!canMove[j][k]) continue;
				if ((i >> k) & 1) continue;
				dp[i + (1 << k)][k] = min(dp[i + (1 << k)][k], dp[i][j] + ecost[j][k]);
			}
		}
	}
	
	int ans = INF;
	for (j = 0; j < n; j++) {
		if (dp[(1 << n) - 1][j] < INF && canMove[j][0]) {
			ans = min(ans, dp[(1 << n) - 1][j] + ecost[j][0]);
		}
	}
	if (ans >= INF) cout << -1 << endl;
	else cout << ans << endl;
	return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

For a given weighted directed graph G(V, E), find the distance of the shortest route that meets the following criteria:

* It is a closed cycle where it ends at the same point it starts.
* It visits each vertex exactly once.

Constraints

* 2 ≤ |V| ≤ 15
* 0 ≤ di ≤ 1,000
* There are no multiedge

Input


|V| |E|
s0 t0 d0
s1 t1 d1
:
s|E|-1 t|E|-1 d|E|-1


|V| is the number of vertices and |E| is the number of edges in the graph. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively.

si and ti represent source and target vertices of i-th edge (directed) and di represents the distance between si and ti (the i-th edge).

Output

Print the shortest distance in a line. If there is no solution, print -1.

Examples

Input

4 6
0 1 2
1 2 3
1 3 9
2 0 1
2 3 6
3 2 4


Output

16


Input

3 3
0 1 1
1 2 1
0 2 1


Output

-1

### Response

```cpp
//ループverです。移動後はvisitedが増えるので, visited昇順にループを回せば良いです。
#include <iostream>
#include <algorithm>
using namespace std;

int INF = 1000000007;
int n, m;
bool canMove[15][15] = {false};
int ecost[15][15];
int dp[1 << 15][15];	//dp[visited][pos] = これまでの最小コスト（visitedの下位i(>=0)ビット目が1⇔頂点iを訪問済み, pos…現在地.）

int main() {
	int i, j, k;
	
	cin >> n >> m;
	for (i = 0; i < m; i++) {
		int s, t, d;
		cin >> s >> t >> d;
		ecost[s][t] = d;
		canMove[s][t] = true;
	}
	
	for (i = 0; i < (1 << n); i++) for (j = 0; j < n; j++) dp[i][j] = INF;
	dp[1][0] = 0;
	for (i = 1; i < (1 << n) - 1; i++) {
		for (j = 0; j < n; j++) {
			if (dp[i][j] >= INF) continue;	//この行は無くてもよい
			for (k = 0; k < n; k++) {
				if (!canMove[j][k]) continue;
				if ((i >> k) & 1) continue;
				dp[i + (1 << k)][k] = min(dp[i + (1 << k)][k], dp[i][j] + ecost[j][k]);
			}
		}
	}
	
	int ans = INF;
	for (j = 0; j < n; j++) {
		if (dp[(1 << n) - 1][j] < INF && canMove[j][0]) {
			ans = min(ans, dp[(1 << n) - 1][j] + ecost[j][0]);
		}
	}
	if (ans >= INF) cout << -1 << endl;
	else cout << ans << endl;
	return 0;
}

```