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;


#define rep(i, n) for(int i = 0; i < (n); i++)


vector<int> ans = {};
int l = 9999;

void solve() {
	int n, m; cin >> n >> m;
	vector<int> b[1000];
	rep(i, m) {
		int A, B; cin >> A >> B;
		b[A - 1].push_back(B - 1);
	}
	// distance, previous
	pair<int, int> d[1001];
	rep(i, n) {
		rep(p, 1001) d[p] = make_pair(9999, -1);
		// previous, present
		queue<pair<int, int>> q; q.push(make_pair(1000, i));
		while (!q.empty()) {
			pair<int, int> p = q.front(); q.pop();
			if (p.second == i && d[p.first].first < l) {
				l = d[p.first].first;
				ans.clear();
				for (int t = p.first; t != 1000; t = d[t].second) {
					ans.push_back(t + 1);
				}
			}

			if (d[p.second].first != 9999) continue;
			rep(j, b[p.second].size()) {
				q.push(make_pair(p.second, b[p.second][j]));
			}
			if (p.first == 1000) d[p.second].first = 0;
			else d[p.second].first = d[p.first].first + 1;
			d[p.second].second = p.first;
		}
	}

	if (l == 9999) { cout << -1 << endl; return; }

	cout << ans.size() << endl;
	rep(i, ans.size()) {
		cout << ans[i] << endl;
	}
}

int main() {


	solve();


	return 0;
}