#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
int n, m, a, b, s, cant[1005], used[1005], w;
vector<int> G[1005];
bool dfs(int t){
	bool f = false;
	used[t] = true;
	for(int u: G[t]){
		cant[u]++;
		if(used[u]){
			if(u != s){
				used[t] = false;
				for(int u: G[t]) cant[u]--;
				return false;
			}else f = true;
		}
	}
	if(f) return true;
	for(int u: G[t]) if(cant[u] == 1 && dfs(u)) return true;
	for(int u: G[t]) cant[u]--;
	used[t] = false;
	return false;
}
int main(){
	scanf("%d%d", &n, &m);
	rep(i,m){
		scanf("%d%d", &a, &b);
		G[a-1].push_back(b-1);
	}
	rep(i,n){
		fill(cant, cant+n, 0);
		fill(used, used+n, false);
		if(dfs(s = i)){
			rep(j,n) if(used[j]) w++;
			printf("%d\n", w);
			rep(j,n) if(used[j]) printf("%d\n", j+1);
			return 0;
		}
	}
	printf("%d\n", -1);
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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;
#define rep(i,n) for(int i=0; i<(n); i++)
int n, m, a, b, s, cant[1005], used[1005], w;
vector<int> G[1005];
bool dfs(int t){
	bool f = false;
	used[t] = true;
	for(int u: G[t]){
		cant[u]++;
		if(used[u]){
			if(u != s){
				used[t] = false;
				for(int u: G[t]) cant[u]--;
				return false;
			}else f = true;
		}
	}
	if(f) return true;
	for(int u: G[t]) if(cant[u] == 1 && dfs(u)) return true;
	for(int u: G[t]) cant[u]--;
	used[t] = false;
	return false;
}
int main(){
	scanf("%d%d", &n, &m);
	rep(i,m){
		scanf("%d%d", &a, &b);
		G[a-1].push_back(b-1);
	}
	rep(i,n){
		fill(cant, cant+n, 0);
		fill(used, used+n, false);
		if(dfs(s = i)){
			rep(j,n) if(used[j]) w++;
			printf("%d\n", w);
			rep(j,n) if(used[j]) printf("%d\n", j+1);
			return 0;
		}
	}
	printf("%d\n", -1);
}
```