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 <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

int N, M, A[1 << 18], B[1 << 18];
pair<int, int> dist[1 << 18];
vector<int> X[1 << 18], Y;

int main() {
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		cin >> A[i] >> B[i];
		X[A[i]].push_back(B[i]);
	}
	for (int i = 0; i < 10000; i++) Y.push_back(i);
	for (int i = 1; i <= N; i++) {
		for (int j = 1; j <= N; j++) dist[j] = make_pair(1 << 30, -1);

		queue<int> Q;
		for (int j : X[i]) {
			dist[j] = make_pair(1, i);
			Q.push(j);
		}
		while (!Q.empty()) {
			int pos = Q.front(); Q.pop();
			for (int j : X[pos]) {
				if (dist[j].first > dist[pos].first + 1) {
					dist[j].first = dist[pos].first + 1;
					dist[j].second = pos;
					Q.push(j);
				}
			}
		}

		if (dist[i].first != (1 << 30)) {
			vector<int> R; int cx = i;
			while (true) {
				cx = dist[cx].second;
				R.push_back(cx);
				if (cx == i) break;
			}
			reverse(R.begin(), R.end());
			if (Y.size() > R.size()) Y = R;
		}
	}
	if (Y.size() == 10000) cout << "-1" << endl;
	else {
		cout << Y.size() << endl;
		for (int i = 0; i < Y.size(); i++) cout << Y[i] << endl;
	}
	return 0;
}