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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

const int INF=1<<29;

int main(){
	int n,m; cin>>n>>m;
	vector<int> G[1000];
	rep(i,m){
		int u,v; cin>>u>>v; u--; v--;
		G[u].emplace_back(v);
	}

	vector<int> ans;
	rep(u,n){
		int d[1000],pre[1000];
		bool vis[1000]={};
		queue<int> Q; Q.push(u);
		rep(v,n) d[v]=INF;
		d[u]=0;
		vis[u]=true;
		while(!Q.empty()){
			int v=Q.front(); Q.pop();
			for(int w:G[v]){
				if(!vis[w]){
					vis[w]=true;
					d[w]=d[v]+1;
					pre[w]=v;
					Q.push(w);
				}
				else if(w==u){
					vector<int> tmp;
					tmp.emplace_back(v);
					int z=pre[v];
					while(z!=u) tmp.emplace_back(z), z=pre[z];
					tmp.emplace_back(u);
					if(ans.empty() || tmp.size()<ans.size()) ans=tmp;
				}
			}
		}
	}

	if(ans.empty()) cout<<-1<<endl;
	else{
		cout<<ans.size()<<endl;
		for(int x:ans) cout<<x+1<<endl;
	}

	return 0;
}