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

Solution (in CPP):

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


int main() {
	int N, M;
	cin >> N >> M;
 
	vector<vector<int>> adj(N);
	while (M--) {
		int a, b;
		cin >> a >> b;
		a--, b--;
		adj[a].push_back(b);
	}
	
	vector<int> res;
    for (int s = 0; s < N; s++) {  // Loop for all vertices; s as the starting vertex.
		vector<int> pred(N, -1);
		queue<int> q;
		q.push(s);
 
		while (q.size()) {
			int i = q.front();
			q.pop();
			for (int j: adj[i]) {
				if (j == s) {
					vector<int> cur = {i};
					while (i != s) i = pred[i], cur.push_back(i);
					if (res.size() == 0 || cur.size() < res.size()) res = cur;
					while (q.size()) q.pop();
					break;
				}
				if (pred[j] != -1) continue;
				pred[j] = i;
				q.push(j);
			}
		}
	}
 
	if (res.size()) {
		cout << res.size() << endl;
		for (int x: res) cout << x + 1 << endl;
	} else {
		cout << -1 << endl;
	}

    return 0;
}