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

const int MAXN = 1e3 + 5;

int N, M;
vector <int> adj[MAXN];
int dist[MAXN], dad[MAXN];

void load() {
	scanf("%d%d", &N, &M);
	while (M--) {
		int a, b;
		scanf("%d%d", &a, &b);
		adj[a].push_back(b);
	}
}

int bfs(int x, int y) {
	memset(dist, -1, sizeof dist);
	dist[y] = 0;
	queue <int> Q;
	Q.push(y);
	while (!Q.empty()) {
		int curr = Q.front();
		Q.pop();
		for (auto it : adj[curr])
			if (dist[it] == -1) {
				dist[it] = dist[curr] + 1;
				dad[it] = curr;
				Q.push(it);
			}
	}
	return dist[x];
}

void solve() {
	int mini = N, u = 0, v = 0;
	for (int i = 1; i <= N; i++)
		for (auto it : adj[i]) {
			int tmp = bfs(i, it);
			if (tmp != -1 && tmp < mini) {
				 mini = tmp;
				 u = i;
				 v = it;
			}
		}
	
	if (mini == N) {
		puts("-1");
		return;
	}
	
	bfs(u, v);
	
	printf("%d\n%d\n", dist[u] + 1, v);
	for (; u != v; u = dad[u])
		printf("%d\n", u);
}

int main() {
	load();
	solve();
	return 0;
}
### Prompt

Create a solution in cpp for 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 MAXN = 1e3 + 5;

int N, M;
vector <int> adj[MAXN];
int dist[MAXN], dad[MAXN];

void load() {
	scanf("%d%d", &N, &M);
	while (M--) {
		int a, b;
		scanf("%d%d", &a, &b);
		adj[a].push_back(b);
	}
}

int bfs(int x, int y) {
	memset(dist, -1, sizeof dist);
	dist[y] = 0;
	queue <int> Q;
	Q.push(y);
	while (!Q.empty()) {
		int curr = Q.front();
		Q.pop();
		for (auto it : adj[curr])
			if (dist[it] == -1) {
				dist[it] = dist[curr] + 1;
				dad[it] = curr;
				Q.push(it);
			}
	}
	return dist[x];
}

void solve() {
	int mini = N, u = 0, v = 0;
	for (int i = 1; i <= N; i++)
		for (auto it : adj[i]) {
			int tmp = bfs(i, it);
			if (tmp != -1 && tmp < mini) {
				 mini = tmp;
				 u = i;
				 v = it;
			}
		}
	
	if (mini == N) {
		puts("-1");
		return;
	}
	
	bfs(u, v);
	
	printf("%d\n%d\n", dist[u] + 1, v);
	for (; u != v; u = dad[u])
		printf("%d\n", u);
}

int main() {
	load();
	solve();
	return 0;
}
```