#include <bits/stdc++.h>

using namespace std;

const int N = 1e3 + 3;

int n, m;
vector <int> g[N];
vector <int> path;

#define pii pair <int, int>
#define mod 10000007

void solve() {
	cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		int u, v;
		cin >> u >> v;
		g[u].push_back(v);
	}
	int minDis = mod;
	for (int i = 1; i <= n; i++) {
		priority_queue <pii, vector <pii >, greater<pii > > pq;
		pq.push({0, i});
		vector <int> dist(n + 1, mod);
		vector <int> par(n + 1, -1);
		while (!pq.empty()) {
			auto top = pq.top();
			pq.pop();
			int u = top.second;
			int c = top.first;
			for (auto v : g[u]) {
				if (dist[v] > c + 1) {
					dist[v] = c + 1;
					par[v] = u;
					pq.push({dist[v], v});
				}
			}
		}
		if (dist[i] < minDis) {
			minDis = dist[i];
			path.clear();
			int idx = i;
			do {
				path.push_back(idx);
				idx = par[idx];
			} while (idx != i);
		}
	}
	if (minDis != mod) {
		cout << (int)path.size() << "\n";
		for (auto el : path) {
			cout << el << "\n";
		}
	} else {
		cout << "-1\n";
	}
}

int main() {
	int t = 1;
	// cin >> t;
	while (t--) {
		solve();
	}
	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;

const int N = 1e3 + 3;

int n, m;
vector <int> g[N];
vector <int> path;

#define pii pair <int, int>
#define mod 10000007

void solve() {
	cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		int u, v;
		cin >> u >> v;
		g[u].push_back(v);
	}
	int minDis = mod;
	for (int i = 1; i <= n; i++) {
		priority_queue <pii, vector <pii >, greater<pii > > pq;
		pq.push({0, i});
		vector <int> dist(n + 1, mod);
		vector <int> par(n + 1, -1);
		while (!pq.empty()) {
			auto top = pq.top();
			pq.pop();
			int u = top.second;
			int c = top.first;
			for (auto v : g[u]) {
				if (dist[v] > c + 1) {
					dist[v] = c + 1;
					par[v] = u;
					pq.push({dist[v], v});
				}
			}
		}
		if (dist[i] < minDis) {
			minDis = dist[i];
			path.clear();
			int idx = i;
			do {
				path.push_back(idx);
				idx = par[idx];
			} while (idx != i);
		}
	}
	if (minDis != mod) {
		cout << (int)path.size() << "\n";
		for (auto el : path) {
			cout << el << "\n";
		}
	} else {
		cout << "-1\n";
	}
}

int main() {
	int t = 1;
	// cin >> t;
	while (t--) {
		solve();
	}
	return 0;
}
```