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;

vector<vector<int>> g;
deque<int> c;
int v[1111];
int f(int u) {
	v[u] = 1;
	for (int w : g[u]) {
		int r = v[w] ? v[w] > 0 ? w : 0 : f(w);
		if (!r) continue;
		if (r > 0)
			c.push_front(u);
		if (r == u) r = -1;
		return r;
	}
	v[u] = -1;
	return 0;
}
int main() {
	int n, m;
	cin >> n >> m;
	g.resize(n + 1);
	set<pair<int, int>> e;
	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		g[a].push_back(b);
		e.emplace(a, b);
	}
	for (int i = 1; i <= n; i++)
		if (!v[i] && f(i))
			goto pn;
	cout << -1 << endl;
	return 0;
pn:
	for (;;) {
		int i, j;
		for (i = 0; i < c.size(); i++) {
			for (j = 0; j < c.size(); j++) {
				if (i == j) continue;
				if (i + 1 == j) continue;
				if (i == c.size() - 1 && !j) continue;
				if (e.find({ c[i],c[j] }) != e.end())
					goto pf;
			}
		}
		break;
	pf:
		if (i < j) c.erase(c.begin() + i + 1, c.begin() + j);
		else c.erase(c.begin() + i + 1, c.end()), c.erase(c.begin(), c.begin() + j);
	}
	cout << c.size() << '\n';
	for (int v : c) cout << v << '\n';
	cout << flush;
	return 0;
}