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 ll long long
#define ld long double

vector<vector<int>> g;
int n, m;
bool used[1000];

vector<int> tps;

bool exist(int s, int t) {
	if (g[s].size() == 0) return false;
	if (lower_bound(g[s].begin(), g[s].end(), t) == g[s].end())
		return false;
	if (*lower_bound(g[s].begin(), g[s].end(), t) == t)
		return true;
	return false;
}

void dfs(int u) {
	if (used[u]) return;
	used[u] = true;
	for (auto& i : g[u]) dfs(i);
	// 帰りがけ順で追加
	tps.push_back(u);
}

void tsort() {
	for (int i = 0; i < n; ++i) dfs(i);
	reverse(tps.begin(), tps.end());
}


signed main() {
	cin >> n >> m;
	g.resize(n);
	for (int i = 0; i < m; ++i) {
		int a, b;
		cin >> a >> b;
		--a; --b;
		g[a].push_back(b);
	}
	for (int i = 0; i < n; ++i)
		sort(g[i].begin(), g[i].end());

	tsort();
	//for (int i : tps) cout << i << " ";
	//cout << endl;

	vector<bool> visited(n, false);
	for (int i = 0; i < tps.size(); ++i) {
		visited[tps[i]] = true;
		for (auto c : g[tps[i]]) {
			if (visited[c]) {
				// TODO
				cout << 0 << endl;
				exit(0);
			}
		}
	}

	cout << -1 << endl;

	return 0;
}