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;
const int maxn = 1010;
vector<int> to[maxn], tree[maxn];
int dist[maxn], par[maxn];
set<int> now;
void dfs(int u, int r){
	now.insert(u);
	bool bad = false, can = false;;
	for(int j : to[u]){
		if(now.count(j)){
			if(j != r){
				bad = true;
			} else {
				can = true;
			}
		}
	}
	if(bad){
		now.erase(u);
		return;
	}
	if(can){
		cout << now.size() << endl;
		for(int v : now) cout << v << endl;
		exit(0);
	}
	for(int  v: tree[u]) dfs(v, r);
	now.erase(u);
}
int main(){
	int n, m; cin >> n >> m;
	for(int i = 0; i < m; i++){
		int a, b; cin >> a >> b;
		to[a].push_back(b);
	}
	for(int i = 1; i <= n; i++){
		memset(dist, -1, sizeof(dist));
		memset(par, -1, sizeof(par));
		for(int j = 0; j <= n; j++) tree[j].clear();
		queue<int> q;
		dist[i] = 0;
		q.push(i);
		while(!q.empty()){
			int u = q.front(); q.pop();
			for(int v : to[u]){
				if(dist[v] == -1){
					dist[v] = dist[u] + 1;
					tree[u].push_back(v);
					q.push(v);
				}
			}
		}
		dfs(i, i);
	}
	cout << -1 << endl;
	return 0;
}