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>
//#include<iostream>
using namespace std;

typedef long long ll;
const int N = 2003;

int n,m;
vector<int> adj[N];
vector<int> ans;

int vstd[N];
vector<int> sol;

bool check(){
	set<int> st;
	int cnt = 0;
	for(int x:ans){
		st.insert(x);
	}
	for(int x:ans){
		for(int y:adj[x]){
			if(st.count(y))cnt++;
		}
	}
	return cnt == (int)st.size();
}

void dfs(int x){
	vstd[x] = 1;
	sol.push_back(x);
	for(int y:adj[x]){
		if((int)ans.size() > 0)return ;
		if(vstd[y] == 2)continue;
		if(vstd[y] == 1){
			int sz = (int)sol.size();
			for(int i=sz-1;i>=0;i--){
				ans.push_back(sol[i]);
				if(sol[i] == y)break;
			}
			if(check()) return;
			else ans.clear();
			/*
			while(sol.back() != y){
				ans.push_back(sol.back());
				sol.pop_back();
			}
			ans.push_back(y);
			return ;
			*/
		}
		else dfs(y);
	}
	if((int)ans.size() > 0)return;
	sol.pop_back();
	vstd[x] = 2;
}

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	
	cin>>n>>m;
	for(int i=1;i<=m;i++){
		int x,y;
		cin>>x>>y;
		adj[x].push_back(y);
	}
	
	for(int i=1;i<=n;i++){
		if(!vstd[i]){
			dfs(i);
			if((int)ans.size() > 0){
				break;
			}
		}
	}

	if((int)ans.size() > 0){
		cout<<(int)ans.size()<<endl;
		for(int x:ans) cout<<x<<endl;
	}
	else{
		cout<<-1;
	}
	
	return 0;
}