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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

namespace{
vector<vector<int>> G;
vector<int> seq;
vector<int> idxs;

void dfs(int u){
	if(idxs[u] >= 0){
		seq.erase(seq.begin(), seq.begin() + idxs[u]);
		throw 0;
	}
	if(idxs[u] == -2){ return; }
	idxs[u] = seq.size();
	seq.push_back(u);
	for(int v : G[u]){ dfs(v); }
	seq.pop_back();
	idxs[u] = -2;
}

void main2(){
	int n, m;
	cin >> n >> m;
	G.resize(n);
	for(int i = 0; i < m; ++i){
		int a, b;
		cin >> a >> b;
		G[a - 1].push_back(b - 1);
	}
	idxs.assign(n, -1);
	try{
		for(int i = 0; i < n; ++i){
			if(idxs[i] == -1){ dfs(i); }
		}
		cout << "-1\n";
		return;
	}
	catch(int){}

	for(int z = 0; z < n; ++z){
		idxs.assign(n, -1);
		for(int i = 0; i < (int)seq.size(); ++i){
			idxs[seq[i]] = i;
		}
		int u = seq[0];
		int mx = 1;
		for(int v : G[u]){ mx = max(mx, idxs[v]); }
		seq.erase(seq.begin(), seq.begin() + mx);
		seq.push_back(u);
	}
	cout << seq.size() << '\n';
	for(int u : seq){ cout << (u + 1) << '\n'; }
}
}

int main(){
	ios::sync_with_stdio(false);
	main2();
}