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 <algorithm>

using namespace std;

int N, M;
int A[2200], B[2200], p[1050];

vector<int> Adj[1005];

int depth[1050], used[1050];

void INIT(void) {
	for (int i = 1; i <= N; i++) {
		depth[i] = 0; used[i] = 0; p[i] = -1;
	}
}

int min_depth;
vector<int> ans;

void update(int K, int h) {

	//cout << "K= " << K << "  h= " << h << endl;
	
	int x = K;
	vector<int> tmp;
	while (1) {
		tmp.push_back(x);
		//cout << x << endl;
		if (x == h) {
			min_depth = depth[K] - depth[h]; ans = tmp;  break;
		}
		x = p[x]; if (x <= 0) { break; }
	}
}

void CALL(int K, int d) {

	used[K] = 1;
	depth[K] = d + 1;

	for (auto h : Adj[K]) {
		if (used[h] == 1) {
			if (depth[K] - depth[h] < min_depth) { update(K, h); }
		}
		else {
			p[h] = K; CALL(h, d + 1);
		}
	}
}

int main(void) {

	cin >> N >> M;

	for (int i = 1; i <= M; i++) {
		cin >> A[i] >> B[i];
		Adj[A[i]].push_back(B[i]);
	}

	min_depth = 9999;

	for (int i = 1; i <= N; i++) {
		INIT();
		p[i] = 0;
		CALL(i, 0);
	}

	if (ans.size() == 0) { cout << -1 << endl; }
	else {
		cout << ans.size() << endl;
		for (int i = ans.size() - 1; i >= 0; i--) {
			cout << ans[i] << endl;
		}
	}

	return 0;
}