#include <bits/stdc++.h>
using namespace std;

#define pb push_back

using ll = long long;

int main() {
    ios::sync_with_stdio(false);cin.tie(nullptr);
    int n, m; cin >> n >> m;
    vector<vector<int>> g(n);
    for (int i = 0; i < m; ++i) {
        int u, v; cin >> u >> v;
        --u, --v;
        g[u].pb(v);
    }
    vector<int> ans;
    for (int u = 0; u < n; ++u) {
        vector<int> par(n, -1);
        queue<int> q;
        for (int v : g[u]) {
            par[v] = u;
            q.push(v);
        }
        while (!q.empty()) {
            int cur = q.front();
            q.pop();
            if (cur == u) {
                vector<int> take;
                int x = par[u];
                while (x != u) {
                    take.pb(x);
                    x = par[x];
                }
                take.pb(u);
                if (ans.size() == 0 || take.size() < ans.size()) ans = take;
                break;
            }
            for (int nxt : g[cur]) {
                if (par[nxt] == -1) {
                    q.push(nxt);
                    par[nxt] = cur;
                }
            }
        }
    }
    if (ans.size()) {
        cout << ans.size() << "\n";
        for (int u : ans) cout << u + 1 << "\n";
    } else {
        cout << -1 << "\n";
    }
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;

#define pb push_back

using ll = long long;

int main() {
    ios::sync_with_stdio(false);cin.tie(nullptr);
    int n, m; cin >> n >> m;
    vector<vector<int>> g(n);
    for (int i = 0; i < m; ++i) {
        int u, v; cin >> u >> v;
        --u, --v;
        g[u].pb(v);
    }
    vector<int> ans;
    for (int u = 0; u < n; ++u) {
        vector<int> par(n, -1);
        queue<int> q;
        for (int v : g[u]) {
            par[v] = u;
            q.push(v);
        }
        while (!q.empty()) {
            int cur = q.front();
            q.pop();
            if (cur == u) {
                vector<int> take;
                int x = par[u];
                while (x != u) {
                    take.pb(x);
                    x = par[x];
                }
                take.pb(u);
                if (ans.size() == 0 || take.size() < ans.size()) ans = take;
                break;
            }
            for (int nxt : g[cur]) {
                if (par[nxt] == -1) {
                    q.push(nxt);
                    par[nxt] = cur;
                }
            }
        }
    }
    if (ans.size()) {
        cout << ans.size() << "\n";
        for (int u : ans) cout << u + 1 << "\n";
    } else {
        cout << -1 << "\n";
    }
}
```