/// However long the night,
///     The dawn will break
/// ICPC next year

#include<bits/stdc++.h>
using namespace std ;
const int N = 1e3+5 ;
int n ,m ,a ,b ;
vector<int> adj[N] ;
vector<int> v ,ans ;
int len = 1e9 ;
int vis[N] ,pos[N] ,vid ;
void dfs(int p,int root,bool work){
    if(work && p==root){
        if(v.size() < len) len = v.size() ,ans = v ;
        return ;
    }
    if(vis[p]==vid) return ;
    vis[p] = vid ;
    pos[p] = v.size() ;
    v.push_back(p);
    for(int ch:adj[p]) dfs(ch,root,1) ;
    v.pop_back();
}
int main(){
    scanf("%d%d",&n,&m);
    while(m--){
        scanf("%d%d",&a,&b);
        adj[a].push_back(b);
    }
    for(int i=1;i<=n;++i){
        ++vid ;
        dfs(i,i,0);
    }
    if(len>n) return cout << -1,0 ;
    cout << len << endl ;
    for(int go:ans)printf("%d\n",go);
    return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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
/// However long the night,
///     The dawn will break
/// ICPC next year

#include<bits/stdc++.h>
using namespace std ;
const int N = 1e3+5 ;
int n ,m ,a ,b ;
vector<int> adj[N] ;
vector<int> v ,ans ;
int len = 1e9 ;
int vis[N] ,pos[N] ,vid ;
void dfs(int p,int root,bool work){
    if(work && p==root){
        if(v.size() < len) len = v.size() ,ans = v ;
        return ;
    }
    if(vis[p]==vid) return ;
    vis[p] = vid ;
    pos[p] = v.size() ;
    v.push_back(p);
    for(int ch:adj[p]) dfs(ch,root,1) ;
    v.pop_back();
}
int main(){
    scanf("%d%d",&n,&m);
    while(m--){
        scanf("%d%d",&a,&b);
        adj[a].push_back(b);
    }
    for(int i=1;i<=n;++i){
        ++vid ;
        dfs(i,i,0);
    }
    if(len>n) return cout << -1,0 ;
    cout << len << endl ;
    for(int go:ans)printf("%d\n",go);
    return 0;
}

```