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

const int INF = 1e9;

int main()
{
	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].push_back(v);
	}
	int dist = INF;
	vector<int> res;
	for (int i = 0; i < N; i++) {
		vector<int> d(N, INF);
		vector<int> pre(N);
		d[i] = 0;
		queue<int> q;
		q.push(i);
		bool ok = false;
		int x = -1;
		while (!q.empty() && !ok) {
			auto v = q.front(); q.pop();
			for (auto to : G[v]) {
				if (to == i) {
					x = v;
					ok = true;
					break;
				}
				if (d[to] == INF) {
					d[to] = d[v] + 1;
					pre[to] = v;
					q.push(to);
				}
			}
		}
		if (ok) {
			vector<int> vs = { i };
			while (x != i) {
				vs.push_back(x);
				x = pre[x];
			}
			sort(vs.begin(), vs.end());
			if (dist > (int)vs.size()) {
				res = vs;
				dist = vs.size();
			}
		}
	}
	if (dist == INF) {
		cout << -1 << endl;
	}
	else {
		cout << dist << endl;
		for (auto v : res) {
			cout << v + 1 << endl;
		}
	}
	return 0;
}

### 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;

const int INF = 1e9;

int main()
{
	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].push_back(v);
	}
	int dist = INF;
	vector<int> res;
	for (int i = 0; i < N; i++) {
		vector<int> d(N, INF);
		vector<int> pre(N);
		d[i] = 0;
		queue<int> q;
		q.push(i);
		bool ok = false;
		int x = -1;
		while (!q.empty() && !ok) {
			auto v = q.front(); q.pop();
			for (auto to : G[v]) {
				if (to == i) {
					x = v;
					ok = true;
					break;
				}
				if (d[to] == INF) {
					d[to] = d[v] + 1;
					pre[to] = v;
					q.push(to);
				}
			}
		}
		if (ok) {
			vector<int> vs = { i };
			while (x != i) {
				vs.push_back(x);
				x = pre[x];
			}
			sort(vs.begin(), vs.end());
			if (dist > (int)vs.size()) {
				res = vs;
				dist = vs.size();
			}
		}
	}
	if (dist == INF) {
		cout << -1 << endl;
	}
	else {
		cout << dist << endl;
		for (auto v : res) {
			cout << v + 1 << endl;
		}
	}
	return 0;
}

```