#include <bits/stdc++.h>
using namespace std;
const int MX = 1e3+2;
vector<int> adj[MX];
int seen[MX], depth[MX];
int prv[MX], st, anslen = 10000;
vector<int> ans;
bool ok;
void dfs(int i) {
	seen[i] = 1;
	for(int a: adj[i]) {
		if(!seen[a]) {
			prv[a] = i;
			depth[a] = depth[i] + 1;
			dfs(a);
		} else if(seen[a] == 1) {
			if(depth[i] - depth[a] < anslen) {
				anslen = depth[i] - depth[a];
				ans = {};
				int cur = i;
				while(cur != a) {
					ans.push_back(cur);
					cur = prv[cur];
				}
				ans.push_back(a);
			}
		}
	}
	seen[i] = 2;
}
int main() {
	ios::sync_with_stdio(0);cin.tie(0);
	int n, m;
	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++) {
		st = i;
		memset(seen, 0, sizeof seen);
		memset(depth, 0, sizeof depth);
		dfs(st);
	}
	if(anslen == 10000)
		return !(cout << -1);
	cout << ans.size() << '\n';
	for(int a: ans)
		cout << a << ' ';
}
### Prompt

In CPP, your task is to solve 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 MX = 1e3+2;
vector<int> adj[MX];
int seen[MX], depth[MX];
int prv[MX], st, anslen = 10000;
vector<int> ans;
bool ok;
void dfs(int i) {
	seen[i] = 1;
	for(int a: adj[i]) {
		if(!seen[a]) {
			prv[a] = i;
			depth[a] = depth[i] + 1;
			dfs(a);
		} else if(seen[a] == 1) {
			if(depth[i] - depth[a] < anslen) {
				anslen = depth[i] - depth[a];
				ans = {};
				int cur = i;
				while(cur != a) {
					ans.push_back(cur);
					cur = prv[cur];
				}
				ans.push_back(a);
			}
		}
	}
	seen[i] = 2;
}
int main() {
	ios::sync_with_stdio(0);cin.tie(0);
	int n, m;
	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++) {
		st = i;
		memset(seen, 0, sizeof seen);
		memset(depth, 0, sizeof depth);
		dfs(st);
	}
	if(anslen == 10000)
		return !(cout << -1);
	cout << ans.size() << '\n';
	for(int a: ans)
		cout << a << ' ';
}
```