#include <bits/stdc++.h>
#define INT long long int
#define oo 9999987654321ll
using namespace std;
INT n, m, p[2010];
bool chk[2010];
vector<INT> G[2010];
void dfs(int x, int z)
{
    chk[x] = true;
    p[z] = x;
    //for(int i = 0 ; i <= z ; i++ ) cout<<p[i]<<" "; cout<<endl;
    for(int i = 0 ; i<G[x].size() ; i++ )
    {
        int y = G[x][i];
        if( !chk[y] )
            dfs(y, z+1);
        else
        {
            set<INT> t;
            t.insert(y);
            for(int j = z ; p[j] != y ; j-- ) t.insert(p[j]);
            INT cnt = 0;
            for(INT q : t)
                for(INT r : G[q] )
                    if( t.find(r) != t.end() ) cnt++;
            if( t.size() == cnt )
            {
                cout<<cnt<<endl;
                for(INT q : t )  cout<<q<<endl;
                exit(EXIT_SUCCESS);
            }
        }
    }
    chk[x] = false;
}
int main()
{
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i = 0 ; i < m ; i++ )
    {
        INT x, y;
        cin>>x>>y;
        G[x].push_back(y);
    }
    for(int i = 1 ; i <= n ; i++ ) dfs(i, 0);
    cout<<-1<<endl;
}

### Prompt

Please formulate a cpp solution to the following problem:

Given is a directed graph G with N vertices and M edges.
The vertices are numbered 1 to N, and the i-th edge is directed from Vertex A_i to Vertex B_i.
It is guaranteed that the graph contains no self-loops or multiple edges.

Determine whether there exists an induced subgraph (see Notes) of G such that the in-degree and out-degree of every vertex are both 1. If the answer is yes, show one such subgraph.
Here the null graph is not considered as a subgraph.

Constraints

* 1 \leq N \leq 1000
* 0 \leq M \leq 2000
* 1 \leq A_i,B_i \leq N
* A_i \neq B_i
* All pairs (A_i, B_i) are distinct.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M
A_1 B_1
A_2 B_2
:
A_M B_M


Output

If there is no induced subgraph of G that satisfies the condition, print `-1`. Otherwise, print an induced subgraph of G that satisfies the condition, in the following format:


K
v_1
v_2
:
v_K


This represents the induced subgraph of G with K vertices whose vertex set is \\{v_1, v_2, \ldots, v_K\\}. (The order of v_1, v_2, \ldots, v_K does not matter.) If there are multiple subgraphs of G that satisfy the condition, printing any of them is accepted.

Examples

Input

4 5
1 2
2 3
2 4
4 1
4 3


Output

3
1
2
4


Input

4 5
1 2
2 3
2 4
1 4
4 3


Output

-1


Input

6 9
1 2
2 3
3 4
4 5
5 6
5 1
5 2
6 1
6 2


Output

4
2
3
4
5

### Response

```cpp
#include <bits/stdc++.h>
#define INT long long int
#define oo 9999987654321ll
using namespace std;
INT n, m, p[2010];
bool chk[2010];
vector<INT> G[2010];
void dfs(int x, int z)
{
    chk[x] = true;
    p[z] = x;
    //for(int i = 0 ; i <= z ; i++ ) cout<<p[i]<<" "; cout<<endl;
    for(int i = 0 ; i<G[x].size() ; i++ )
    {
        int y = G[x][i];
        if( !chk[y] )
            dfs(y, z+1);
        else
        {
            set<INT> t;
            t.insert(y);
            for(int j = z ; p[j] != y ; j-- ) t.insert(p[j]);
            INT cnt = 0;
            for(INT q : t)
                for(INT r : G[q] )
                    if( t.find(r) != t.end() ) cnt++;
            if( t.size() == cnt )
            {
                cout<<cnt<<endl;
                for(INT q : t )  cout<<q<<endl;
                exit(EXIT_SUCCESS);
            }
        }
    }
    chk[x] = false;
}
int main()
{
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i = 0 ; i < m ; i++ )
    {
        INT x, y;
        cin>>x>>y;
        G[x].push_back(y);
    }
    for(int i = 1 ; i <= n ; i++ ) dfs(i, 0);
    cout<<-1<<endl;
}

```