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>to[1005];
int INF = 1e9;
int N;
vector<int> bfs(int s) {
	vector<int>dist(N, INF);
	vector<int>pre(N, -1);
	queue<int>q;
	dist[s] = 0;
	q.push(s);
	while (!q.empty()) {
		int v = q.front();
		q.pop();
		for (int i : to[v]) {
			if (dist[i] != INF) {
				continue;
			}
			pre[i] = v;
			dist[i] = dist[v] + 1;
			q.push(i);
		}
	}
	pair<int, int>best(INF, -1);
	for (int i = 0; i < N; ++i) {
		if (i == s) {
			continue;
		}
		for (int u :to[i]) {
			if (u == s) {
				best = min(best,make_pair(dist[i],i));
			}
		}
	}
	if (best.first == INF) {
		return vector<int>(N + 1);
	}
	int v = best.second;
	vector<int>res;
	while (-1 != v) {
		res.push_back(v);
		v = pre[v];
	}
	return res;
}

int main() {
	int M;
	cin >> N >> M;
	for (int i = 0; i < M; ++i) {
		int a, b;
		cin >> a >> b;
		--a;
		--b;
		to[a].push_back(b);
	}

	vector<int> ans(N + 1);
	for (int i = 0; i < N; ++i) {
		vector<int> now = bfs(i);
		if (now.size() < ans.size()) {
			ans = now;
		}
	}
	if (ans.size() == (N + 1)) {
		cout << -1 << endl;
		return 0;
	}
	cout << ans.size() << endl;
	for (auto i : ans) {
		cout << i + 1 << endl;
	}
	return 0;
}