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;

long long N, M, MIN = 1000000000;
static vector<long long> Fin;
long long Par[1000] = {}, Rank[1000] = {};
bool visited[1000] = {};
static bool Back[1000][1000] = {};
static vector<long long> To[1000];
static vector<long long> V[1000];

void DFS(long long S, long long P, long long R) {
	To[S].push_back(P);
	Rank[P] = R;
	visited[P] = true;
	for (int X : V[P]) {
		if (!visited[X]) {
			Par[X] = P;
			DFS(S, X, R + 1);
		}
	}
}

int main() {
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		long long A, B;
		cin >> A >> B;
		A--, B--;
		V[A].push_back(B);
		Back[B][A] = true;
	}
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) visited[j] = false, Par[j] = 0, Rank[j] = 1000000000;
		DFS(i, i, 0);
		long long MINX = 1000000000;
		vector<long long> ANSX = {};
		for (int X : To[i]) {
			if (Back[i][X] && Rank[X] < MINX) {
				MINX = Rank[X];
				vector<long long> ANS;
				while (X != i) {
					ANS.push_back(X);
					X = Par[X];
				}
				ANS.push_back(i);
				ANSX = ANS;
			}
		}
		if (!ANSX.empty()) {
			if (ANSX.size() < MIN) {
				MIN = ANSX.size();
				Fin = ANSX;
			}
		}
	}
	if (Fin.empty()) cout << -1 << endl;
	else {
		reverse(Fin.begin(), Fin.end());
		cout << Fin.size() << endl;
		for (int P : Fin) cout << P + 1 << endl;
	}
}