#include <bits/stdc++.h>
#define Int int64_t

using namespace std;

int main() {
	int N, M;
	cin >> N >> M;
	vector<vector<int>> g(N);
	for (int i = 0; i < M; ++i) {
		int a, b;
		cin >> a >> b;
		g[a - 1].emplace_back(b - 1);
	}

	vector<int> ans;
	for (int i = 0; i < N; ++i) {
		vector<bool> used(N, false);
		vector<int> prev(N);
		deque<int> que;
		que.emplace_back(i);
		used[i] = true;
		while (que.size()) {
			int u = que.front(); que.pop_front();
			for (auto v : g[u]) {
				if (v == i) {
					vector<int> tmp;
					tmp.emplace_back(v);
					while (u != i) {
						tmp.emplace_back(u);
						u = prev[u];
					}
					reverse(tmp.begin(), tmp.end());
					if (ans.size() == 0 || ans.size() > tmp.size()) {
						ans = tmp;
					}
					break;
				}
				if (used[v]) { continue; }
				used[v] = true;
				prev[v] = u;
				que.emplace_back(v);
			}
		}
	}
	if (ans.size() == 0) { 
		cout << -1 << endl;
	} else {
		cout << ans.size() << endl;
		for (auto v : ans) { cout << v + 1 << endl; }
	}

	return 0;
}

### Prompt

In cpp, your task is to solve 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 <bits/stdc++.h>
#define Int int64_t

using namespace std;

int main() {
	int N, M;
	cin >> N >> M;
	vector<vector<int>> g(N);
	for (int i = 0; i < M; ++i) {
		int a, b;
		cin >> a >> b;
		g[a - 1].emplace_back(b - 1);
	}

	vector<int> ans;
	for (int i = 0; i < N; ++i) {
		vector<bool> used(N, false);
		vector<int> prev(N);
		deque<int> que;
		que.emplace_back(i);
		used[i] = true;
		while (que.size()) {
			int u = que.front(); que.pop_front();
			for (auto v : g[u]) {
				if (v == i) {
					vector<int> tmp;
					tmp.emplace_back(v);
					while (u != i) {
						tmp.emplace_back(u);
						u = prev[u];
					}
					reverse(tmp.begin(), tmp.end());
					if (ans.size() == 0 || ans.size() > tmp.size()) {
						ans = tmp;
					}
					break;
				}
				if (used[v]) { continue; }
				used[v] = true;
				prev[v] = u;
				que.emplace_back(v);
			}
		}
	}
	if (ans.size() == 0) { 
		cout << -1 << endl;
	} else {
		cout << ans.size() << endl;
		for (auto v : ans) { cout << v + 1 << endl; }
	}

	return 0;
}

```