#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>

using ll = long long;

using namespace std;

int n, m;
vector<int> G[1010], ans, trace;
int from[1010];

void bfs(int i) {
    queue<int> q;
    fill(from, from+n, -1);
    q.push(i);
    while (!q.empty()) {
        int f = q.front(); q.pop();
        for (auto g:G[f]) {
            if (g == i) {
                trace = vector<int>();
                trace.push_back(i + 1);
                while (f != i) {
                    trace.push_back(f + 1);
                    f = from[f];
                }
                if (ans.size() == 0 || ans.size() > trace.size()) swap(ans, trace);
                return;
            }
            if (from[g] >= 0) continue;
            q.push(g);
            from[g] = f;
        }
    }
}

int main() {
    cin >> n >> m;
    for (int i=0, a, b; i<m; i++) {
        cin >> a >> b;
        a--; b--;
        G[a].push_back(b);
    }
    for (int i=0; i<n; i++) bfs(i);
    if (ans.size() == 0)
        cout << -1 << '\n';
    else {
        cout << ans.size() << '\n';
        for (auto ai:ans) cout << ai << '\n';
    }
    return 0;
}
### Prompt

Please create a solution in cpp 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 <algorithm>
#include <vector>
#include <queue>
#include <string>

using ll = long long;

using namespace std;

int n, m;
vector<int> G[1010], ans, trace;
int from[1010];

void bfs(int i) {
    queue<int> q;
    fill(from, from+n, -1);
    q.push(i);
    while (!q.empty()) {
        int f = q.front(); q.pop();
        for (auto g:G[f]) {
            if (g == i) {
                trace = vector<int>();
                trace.push_back(i + 1);
                while (f != i) {
                    trace.push_back(f + 1);
                    f = from[f];
                }
                if (ans.size() == 0 || ans.size() > trace.size()) swap(ans, trace);
                return;
            }
            if (from[g] >= 0) continue;
            q.push(g);
            from[g] = f;
        }
    }
}

int main() {
    cin >> n >> m;
    for (int i=0, a, b; i<m; i++) {
        cin >> a >> b;
        a--; b--;
        G[a].push_back(b);
    }
    for (int i=0; i<n; i++) bfs(i);
    if (ans.size() == 0)
        cout << -1 << '\n';
    else {
        cout << ans.size() << '\n';
        for (auto ai:ans) cout << ai << '\n';
    }
    return 0;
}
```