#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;

int n, m, a, b, vt[1005];
vector<int> gr[1005], tmp, ans;

void dfs(int p) {
	tmp.push_back(p);
	vt[p] = 1;
	for(int x:gr[p]) {
		if(x == tmp[0]) {
			if(ans.empty() || tmp.size() < ans.size()) ans = tmp;
		} else if (!vt[x]) {
			dfs(x);
		}
	}
	tmp.pop_back();
}

int main() {
	cin >> n >> m;
	while(m--) {
		cin >> a >> b;
		gr[a].push_back(b);
	}
	for(int i=1; i<=n; i++) {
		memset(vt, 0, sizeof(vt));
		tmp.clear();
		dfs(i);
	}
	if(ans.empty()) {puts("-1"); return 0;}
	cout << ans.size() << endl;
	for(auto x:ans) cout << x << endl;
	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 <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;

int n, m, a, b, vt[1005];
vector<int> gr[1005], tmp, ans;

void dfs(int p) {
	tmp.push_back(p);
	vt[p] = 1;
	for(int x:gr[p]) {
		if(x == tmp[0]) {
			if(ans.empty() || tmp.size() < ans.size()) ans = tmp;
		} else if (!vt[x]) {
			dfs(x);
		}
	}
	tmp.pop_back();
}

int main() {
	cin >> n >> m;
	while(m--) {
		cin >> a >> b;
		gr[a].push_back(b);
	}
	for(int i=1; i<=n; i++) {
		memset(vt, 0, sizeof(vt));
		tmp.clear();
		dfs(i);
	}
	if(ans.empty()) {puts("-1"); return 0;}
	cout << ans.size() << endl;
	for(auto x:ans) cout << x << endl;
	return 0;
}

```