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 MAXN 1005

vector<int> g[MAXN];
vector<vector<int>> cycles;
int par[MAXN];


void dfs(int u) {
	//$ cerr << u << endl;
	for(int v : g[u]) {
		if(par[v] == -1) {
			par[v] = u;
			dfs(v);
		} else {
			int cur = u;
			vector<int> cyc = {u};
			while(cur != v && par[cur] != -2) {
				cyc.push_back(par[cur]);
				cur = par[cur];
			}
			
			if(cyc.back() == v)
				cycles.push_back(cyc);
		}
	}
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	int n, m;
	cin >> n >> m;
	for(int i = 0; i < m; i++) {
		int u, v;
		cin >> u >> v;
		u--, v--;
		g[u].push_back(v);
	}
	
	
	for(int i = 0; i < n; i++) {
		memset(par, -1, sizeof par);
		par[i] = -2;
		dfs(i);
	}
	
	
	for(vector<int> v : cycles) {
		set<int> s(v.begin(), v.end());
		bool ok = true;
		for(int i = 0; i < (int) v.size(); i++) {
			int cnt = 0;
			for(int nei : g[v[i]]) {
				if(s.count(nei)) {
					cnt++;
				}
			}
			if(cnt > 1) {
				ok = false;
			}
			//$ cerr << v[i] << " : nei's = " << cnt << endl;
		}
		
		//$ cerr << endl;
		
		if(ok) {
			cout << v.size() << '\n';
			for(int i = 0; i < (int) v.size(); i++) {
				cout << v[i]+1 << '\n';
			}
			return 0;
		}
	}
	
	cout << -1 << endl;
}