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<int> edge[1010];
set<int> ans;

void bfs(int start, int goal) {
	bool used[1010];
	int prev[1010];
	for(int i = 0; i < 1010; ++i) {
		used[i] = false;
		prev[i] = -1;
	}
	queue<int> que;
	que.push(start);
	used[start] = true;
	prev[start] = goal;
	while(que.size()) {
		int from = que.front();
		que.pop();
		for(int to : edge[from]) {
			if(not used[to]) {
				used[to] = true;
				prev[to] = from;
				que.push(to);
			}
		}
	}
	set<int> tmp;
	int now = goal;
	while(now != start) {
		if(now == -1) return;
		tmp.insert(now);
		now = prev[now];
	}
	tmp.insert(start);
	if(tmp.size() < ans.size()) {
		ans = tmp;
	}
}

int main() {
	int n, m; cin >> n >> m;
	int s[m], t[m];
	for(int i = 0; i < m; ++i) {
		cin >> s[i] >> t[i];
		edge[--s[i]].push_back(--t[i]);
	}
	for(int i = 0; i <= n; ++i) {
		ans.insert(i);
	}
	for(int i = 0; i < m; ++i) {
		bfs(t[i], s[i]);
	}
	if(ans.size() == n + 1) {
		cout << -1 << '\n';
		return 0;
	}
	cout << ans.size() << '\n';
	for(int i : ans) {
		cout << i + 1 << '\n';
	}
	return 0;
}