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

bool g[1000][1000];
bool visited[1000];
vector<int> res;

void find_cycle(int n, int start, vector<int>& path) {
	visited[start] = true;
	path.push_back(start);
	
	for (int dest = 0; dest < n; ++dest) {
		if (g[start][dest] && dest == path[0] && path.size() > 1 && (res.empty() || path.size() < res.size())) {
			res = path;
		} else if (g[start][dest] && !visited[dest]) {
			find_cycle(n, dest, path);
		}
	}

	path.pop_back();
}

int main() {
	memset(g, false, sizeof(g));

	int n, m;
	cin >> n >> m;

	for (int i = 0; i < m; ++i) {
		int a, b;
		cin >> a >> b;
		g[a - 1][b - 1] = true;
	}

	for (int i = 0; i < n; ++i) {
		memset(visited, false, sizeof(visited));
		vector<int> p;
		find_cycle(n, i, p);
	}

	if (res.empty()) {
		cout << -1;
		return 0;
	}

	cout << res.size() << endl;
	for (int c : res) {
		cout << c + 1 << endl;
	}
	
	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 <bits/stdc++.h>
using namespace std;

bool g[1000][1000];
bool visited[1000];
vector<int> res;

void find_cycle(int n, int start, vector<int>& path) {
	visited[start] = true;
	path.push_back(start);
	
	for (int dest = 0; dest < n; ++dest) {
		if (g[start][dest] && dest == path[0] && path.size() > 1 && (res.empty() || path.size() < res.size())) {
			res = path;
		} else if (g[start][dest] && !visited[dest]) {
			find_cycle(n, dest, path);
		}
	}

	path.pop_back();
}

int main() {
	memset(g, false, sizeof(g));

	int n, m;
	cin >> n >> m;

	for (int i = 0; i < m; ++i) {
		int a, b;
		cin >> a >> b;
		g[a - 1][b - 1] = true;
	}

	for (int i = 0; i < n; ++i) {
		memset(visited, false, sizeof(visited));
		vector<int> p;
		find_cycle(n, i, p);
	}

	if (res.empty()) {
		cout << -1;
		return 0;
	}

	cout << res.size() << endl;
	for (int c : res) {
		cout << c + 1 << endl;
	}
	
	return 0;
}

```