#include <bits/stdc++.h>
#define X first
#define Y second
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define ini(x, y) memset(x, y, sizeof(x))
#define endl '\n'
#define fastio cin.sync_with_stdio(false); cin.tie(nullptr)
using namespace std;

using ll = long long;
using pii = pair<int, int>;
const int dx[] = { -1, 0, 1, 0, -1, 1, 1, -1 };
const int dy[] = { 0, 1, 0, -1, 1, 1, -1, -1 };

vector<int> adj[1000];

int main() {
	fastio;
	int N, M;
	cin >> N >> M;

	for (int a, b, i = 0; i < M; ++i) {
		cin >> a >> b;
		adj[--a].push_back(--b);
	}

	vector<int> ans;
	for (int i = 0; i < N; ++i) {
		int wif[1000];
		ini(wif, -1);

		queue<int> Q;
		for (int nxt : adj[i]) {
			Q.push(nxt);
			wif[nxt] = i;
		}

		while (!Q.empty()) {
			int now = Q.front(); Q.pop();

			if (now == i) {
				vector<int> tmp = { i };
				for (int j = wif[i]; j != i; j = wif[j]) tmp.push_back(j);

				if (ans.empty() || sz(ans) > sz(tmp)) ans = tmp;
				break;
			}

			for (int nxt : adj[now]) if (wif[nxt] == -1) {
				wif[nxt] = now;
				Q.push(nxt);
			}
		}
	}

	if (ans.empty()) return cout << -1, 0;

	cout << sz(ans) << endl;
	for (int i : ans) cout << i + 1 << endl;

	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
#include <bits/stdc++.h>
#define X first
#define Y second
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define ini(x, y) memset(x, y, sizeof(x))
#define endl '\n'
#define fastio cin.sync_with_stdio(false); cin.tie(nullptr)
using namespace std;

using ll = long long;
using pii = pair<int, int>;
const int dx[] = { -1, 0, 1, 0, -1, 1, 1, -1 };
const int dy[] = { 0, 1, 0, -1, 1, 1, -1, -1 };

vector<int> adj[1000];

int main() {
	fastio;
	int N, M;
	cin >> N >> M;

	for (int a, b, i = 0; i < M; ++i) {
		cin >> a >> b;
		adj[--a].push_back(--b);
	}

	vector<int> ans;
	for (int i = 0; i < N; ++i) {
		int wif[1000];
		ini(wif, -1);

		queue<int> Q;
		for (int nxt : adj[i]) {
			Q.push(nxt);
			wif[nxt] = i;
		}

		while (!Q.empty()) {
			int now = Q.front(); Q.pop();

			if (now == i) {
				vector<int> tmp = { i };
				for (int j = wif[i]; j != i; j = wif[j]) tmp.push_back(j);

				if (ans.empty() || sz(ans) > sz(tmp)) ans = tmp;
				break;
			}

			for (int nxt : adj[now]) if (wif[nxt] == -1) {
				wif[nxt] = now;
				Q.push(nxt);
			}
		}
	}

	if (ans.empty()) return cout << -1, 0;

	cout << sz(ans) << endl;
	for (int i : ans) cout << i + 1 << endl;

	return 0;
}
```