#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
typedef long long int ll;

vector<int> g[1001];
int c[1010][1010];
vector<int> ans;

void dfs(int s,int cnt,int f){
    for(int t:g[s]){
        if(c[f][t]>cnt){
            c[f][t]=cnt;
            dfs(t,cnt+1,f);
        }
    }
}

void dfs2(int s,int f){
    ans.push_back(s);
    int to=-1;
    int mi=1e6;
    for(int t:g[s]){
        if(t==f)return;
        if(mi>c[t][f]){
            mi=c[t][f];
            to=t;
        }
    }
    dfs2(to,f);
}

int main(){
    int n,m; cin >> n >> m;
    for(int i=0;i<m;i++){
        int a,b; cin >> a >> b;
        a--; b--;
        g[a].push_back(b);
    }
    for(int i=0;i<1001;i++){
        for(int j=0;j<1001;j++){
            c[i][j]=1e9;
        }
    }
    for(int i=0;i<n;i++){
        dfs(i,1,i);
    }
    int mi=1e9; int s=-1;
    for(int i=0;i<n;i++){
        if(mi>c[i][i]){
            mi=c[i][i];
            s=i;
        }
    }
    if(s==-1){
        cout << -1 << endl;
        return 0;
    }
    dfs2(s,s);
    cout << ans.size() << endl;
    for(int i=0;i<ans.size();i++){
        cout << ans[i]+1 << endl;
    }
}
### Prompt

Your task is to create 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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
typedef long long int ll;

vector<int> g[1001];
int c[1010][1010];
vector<int> ans;

void dfs(int s,int cnt,int f){
    for(int t:g[s]){
        if(c[f][t]>cnt){
            c[f][t]=cnt;
            dfs(t,cnt+1,f);
        }
    }
}

void dfs2(int s,int f){
    ans.push_back(s);
    int to=-1;
    int mi=1e6;
    for(int t:g[s]){
        if(t==f)return;
        if(mi>c[t][f]){
            mi=c[t][f];
            to=t;
        }
    }
    dfs2(to,f);
}

int main(){
    int n,m; cin >> n >> m;
    for(int i=0;i<m;i++){
        int a,b; cin >> a >> b;
        a--; b--;
        g[a].push_back(b);
    }
    for(int i=0;i<1001;i++){
        for(int j=0;j<1001;j++){
            c[i][j]=1e9;
        }
    }
    for(int i=0;i<n;i++){
        dfs(i,1,i);
    }
    int mi=1e9; int s=-1;
    for(int i=0;i<n;i++){
        if(mi>c[i][i]){
            mi=c[i][i];
            s=i;
        }
    }
    if(s==-1){
        cout << -1 << endl;
        return 0;
    }
    dfs2(s,s);
    cout << ans.size() << endl;
    for(int i=0;i<ans.size();i++){
        cout << ans[i]+1 << endl;
    }
}
```