#include <bits/stdc++.h>

using namespace std;

vector<int> graf[1005], cykl;

int poprzednik[1005], odwiedzony[1005];

void dfs(int x){
    odwiedzony[x] = 1;
    for (int i : graf[x]){
        if (odwiedzony[i] == 1){
            vector<int> nowy = {i};
            int pom = x;
            while (pom != i){
                nowy.push_back(pom);
                pom = poprzednik[pom];
            }
            if (!cykl.size() || nowy.size() < cykl.size())
                cykl = nowy;
        }
        if (!odwiedzony[i]){
            poprzednik[i] = x;
            dfs(i);
        }
    }
    odwiedzony[x] = 2;
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, m, a, b;
    cin >> n >> m;
    for (int i = 0; i < m; i++){
        cin >> a >> b;
        graf[a].push_back(b);
    }
    for (int i = 1; i <= n; i++)
        random_shuffle(graf[i].begin(), graf[i].end());
    for (int i = 1; i <= n; i++)
        if (!odwiedzony[i])
            dfs(i);
    if (!cykl.size()){
        cout << "-1\n";
        return 0;
    }
    cout << cykl.size() << "\n";
    for (int i : cykl)
        cout << i << "\n";
    return 0;
}

### Prompt

Develop a solution in CPP 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
#include <bits/stdc++.h>

using namespace std;

vector<int> graf[1005], cykl;

int poprzednik[1005], odwiedzony[1005];

void dfs(int x){
    odwiedzony[x] = 1;
    for (int i : graf[x]){
        if (odwiedzony[i] == 1){
            vector<int> nowy = {i};
            int pom = x;
            while (pom != i){
                nowy.push_back(pom);
                pom = poprzednik[pom];
            }
            if (!cykl.size() || nowy.size() < cykl.size())
                cykl = nowy;
        }
        if (!odwiedzony[i]){
            poprzednik[i] = x;
            dfs(i);
        }
    }
    odwiedzony[x] = 2;
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, m, a, b;
    cin >> n >> m;
    for (int i = 0; i < m; i++){
        cin >> a >> b;
        graf[a].push_back(b);
    }
    for (int i = 1; i <= n; i++)
        random_shuffle(graf[i].begin(), graf[i].end());
    for (int i = 1; i <= n; i++)
        if (!odwiedzony[i])
            dfs(i);
    if (!cykl.size()){
        cout << "-1\n";
        return 0;
    }
    cout << cykl.size() << "\n";
    for (int i : cykl)
        cout << i << "\n";
    return 0;
}

```