#include<algorithm>
#include<vector>
#include<limits>
#include<cstdio>
#include<cstring>
#define rep(i,a) for(int i=0;i<(a);++i)
#define clr(a,v) memset((a),(v),sizeof(a))

const int MAX_V = 15, INF = std::numeric_limits<int>::max()>>2;

int V, E;
std::vector<std::vector<int> > d( MAX_V, std::vector<int>( MAX_V, INF ) );
int dp[1<<MAX_V+1][MAX_V];

int sv;

int rec( int S, int v )
{
    if( S == (1<<V)-1 && v == sv )
        return dp[S][v] = 0;

    if( ~dp[S][v] )
        return dp[S][v];

    int ret = INF;

    rep( u, V ) if( !(S>>u&1) )
        ret = std::min( ret, rec( S|1<<u, u )+d[v][u] );

    return dp[S][v] = ret;
}

int main()
{
    scanf( "%d%d", &V, &E );
    rep( i, E )
    {
        int s, t;
        scanf( "%d%d", &s, &t );
        scanf( "%d", &d[s][t] );
    }

    int ans = INF;
    rep( v, V )
    {
        clr( dp, -1 );
        sv = v;
        ans = std::min( ans, rec( 0, v ) );
    }

    printf( "%d\n", ans==INF?-1:ans );

    return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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<algorithm>
#include<vector>
#include<limits>
#include<cstdio>
#include<cstring>
#define rep(i,a) for(int i=0;i<(a);++i)
#define clr(a,v) memset((a),(v),sizeof(a))

const int MAX_V = 15, INF = std::numeric_limits<int>::max()>>2;

int V, E;
std::vector<std::vector<int> > d( MAX_V, std::vector<int>( MAX_V, INF ) );
int dp[1<<MAX_V+1][MAX_V];

int sv;

int rec( int S, int v )
{
    if( S == (1<<V)-1 && v == sv )
        return dp[S][v] = 0;

    if( ~dp[S][v] )
        return dp[S][v];

    int ret = INF;

    rep( u, V ) if( !(S>>u&1) )
        ret = std::min( ret, rec( S|1<<u, u )+d[v][u] );

    return dp[S][v] = ret;
}

int main()
{
    scanf( "%d%d", &V, &E );
    rep( i, E )
    {
        int s, t;
        scanf( "%d%d", &s, &t );
        scanf( "%d", &d[s][t] );
    }

    int ans = INF;
    rep( v, V )
    {
        clr( dp, -1 );
        sv = v;
        ans = std::min( ans, rec( 0, v ) );
    }

    printf( "%d\n", ans==INF?-1:ans );

    return 0;
}
```