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

int n, m, pre[1001];
bool vis[1001];
vector<int> adj[1001];

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin >> n >> m;
	for(int i = 0; i < m; i++){
		int a, b;
		cin >> a >> b;
		adj[a].push_back(b);
	}
	for(int i = 1; i <= n; i++){
		memset(vis, 0, sizeof(vis));
		memset(pre, 0, sizeof(pre));
		stack<pair<int, int>> que;
		que.push({i, 0});
		while(!que.empty()){
			pair<int, int> vtx = que.top();
			que.pop();
			vis[vtx.first] = true;
			pre[vtx.first] = vtx.second;
			for(auto j : adj[vtx.first])
				if(j == i){
					set<int> chn = {i};
					int cur = vtx.first;
					while(cur != i){
						chn.insert(cur);
						cur = pre[cur];
					}
					bool val = true;
					for(auto k : chn){
						int cnt = 0;
						for(auto l : adj[k])
							if(chn.find(l) != chn.end()) cnt++;
						if(cnt > 1) val = false;
					}
					if(val){
						cout << chn.size() << "\n";
						for(auto k : chn) cout << k << "\n";
						return 0;
					}
				}else if(!vis[j]) que.push({j, vtx.first});
		}
	}
	cout << "-1\n";
	return 0;
}
### Prompt

Your challenge is to write a Cpp solution 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;

int n, m, pre[1001];
bool vis[1001];
vector<int> adj[1001];

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin >> n >> m;
	for(int i = 0; i < m; i++){
		int a, b;
		cin >> a >> b;
		adj[a].push_back(b);
	}
	for(int i = 1; i <= n; i++){
		memset(vis, 0, sizeof(vis));
		memset(pre, 0, sizeof(pre));
		stack<pair<int, int>> que;
		que.push({i, 0});
		while(!que.empty()){
			pair<int, int> vtx = que.top();
			que.pop();
			vis[vtx.first] = true;
			pre[vtx.first] = vtx.second;
			for(auto j : adj[vtx.first])
				if(j == i){
					set<int> chn = {i};
					int cur = vtx.first;
					while(cur != i){
						chn.insert(cur);
						cur = pre[cur];
					}
					bool val = true;
					for(auto k : chn){
						int cnt = 0;
						for(auto l : adj[k])
							if(chn.find(l) != chn.end()) cnt++;
						if(cnt > 1) val = false;
					}
					if(val){
						cout << chn.size() << "\n";
						for(auto k : chn) cout << k << "\n";
						return 0;
					}
				}else if(!vis[j]) que.push({j, vtx.first});
		}
	}
	cout << "-1\n";
	return 0;
}
```