#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define P pair<ll,ll>
#define FOR(I,A,B) for(ll I = (A); I < (B); ++I)
#define FORR(I,A,B) for(ll I = ((B)-1); I >= (A); --I)
#define SORT(x) (sort(x.begin(),x.end())) // 0 2 2 3 4 5 8 9
#define REV(x) (reverse(x.begin(),x.end())) //reverse
ll gcd(ll a,ll b){if(a<b)swap(a,b);if(a%b==0)return b;else return gcd(b,a%b);}
ll lcm(ll a,ll b){ll c=gcd(a,b);return ((a/c)*(b/c)*c);}//saisyo kobaisu
#define NEXTP(x) next_permutation(x.begin(),x.end())
const ll INF=1e18+7;
const ll MOD=1e9+7;

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	ll n,m;
	cin >> n >> m;
	ll dp[(1<<n)][n];
	ll d[n][n];
	FOR(i,0,n)FOR(j,0,n)d[i][j]=INF;
	FOR(i,0,m){
		ll a,b,c;
		cin >> a >> b >> c;
		d[a][b]  = c;
	}
	FOR(i,0,(1<<n))
		FOR(j,0,n)
			dp[i][j]=INF;
	dp[(1<<n)-1][0] = 0;
	FORR(S,0,(1<<n)-1){
		FOR(v,0,n){
			FOR(u,0,n){
				if(!(S>>u&1))
					dp[S][v] = min(dp[S][v],dp[S|1<<u][u]+d[u][v]);
			}
		}
	}
	if(dp[0][0]<INF)cout << dp[0][0] << endl;
	else cout << "-1" << endl;
}


### Prompt

Your task is to create 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
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define P pair<ll,ll>
#define FOR(I,A,B) for(ll I = (A); I < (B); ++I)
#define FORR(I,A,B) for(ll I = ((B)-1); I >= (A); --I)
#define SORT(x) (sort(x.begin(),x.end())) // 0 2 2 3 4 5 8 9
#define REV(x) (reverse(x.begin(),x.end())) //reverse
ll gcd(ll a,ll b){if(a<b)swap(a,b);if(a%b==0)return b;else return gcd(b,a%b);}
ll lcm(ll a,ll b){ll c=gcd(a,b);return ((a/c)*(b/c)*c);}//saisyo kobaisu
#define NEXTP(x) next_permutation(x.begin(),x.end())
const ll INF=1e18+7;
const ll MOD=1e9+7;

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	ll n,m;
	cin >> n >> m;
	ll dp[(1<<n)][n];
	ll d[n][n];
	FOR(i,0,n)FOR(j,0,n)d[i][j]=INF;
	FOR(i,0,m){
		ll a,b,c;
		cin >> a >> b >> c;
		d[a][b]  = c;
	}
	FOR(i,0,(1<<n))
		FOR(j,0,n)
			dp[i][j]=INF;
	dp[(1<<n)-1][0] = 0;
	FORR(S,0,(1<<n)-1){
		FOR(v,0,n){
			FOR(u,0,n){
				if(!(S>>u&1))
					dp[S][v] = min(dp[S][v],dp[S|1<<u][u]+d[u][v]);
			}
		}
	}
	if(dp[0][0]<INF)cout << dp[0][0] << endl;
	else cout << "-1" << endl;
}


```