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() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	
	int n, m;
	cin >> n >> m;

	vector<vector<int> > G;
	vector<vector<int> > rev;
	G.resize(n);
	rev.resize(n);

	for(int i = 0; i < m; ++i) {
		int a, b;
		cin >> a >> b;
		--a;
		--b;
		G[a].emplace_back(b);
		rev[b].emplace_back(a);
	}

	vector<int> vis;
	
	set<int> st;

	function<void(int, int, int)> dfs = [&] (int u, int s, int f) {
		vis[u] = 1;
		st.emplace(u);
		for(auto v : G[u]) {
			if(v != s && st.count(v)) {
				st.erase(u);
				return;
			}
		}
		for(auto v : rev[u]) {
			if(v != f && st.count(v)) {
				st.erase(u);
				return;
			}
		}
		for(auto v : G[u]) {
			if(v == s) {
				cout << st.size() << '\n';
				for(auto x : st) {
					cout << x + 1 << '\n';
				}
				exit(0);
			}
		}
		for(auto v : G[u]) {
			if(vis[v] == 0) {
				dfs(v, s, u);
			}
		}
		st.erase(u);
	};

	for(int i = 0; i < n; ++i) {
		st.clear();
		vis.assign(n, 0);
		dfs(i, i, -1);
	}
	
	cout << -1 << '\n';

	return 0;
}