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;

int fa[2005];
int b[2005],a[2005];

int find(int x,int &cnt,int y)
{
	b[cnt++] = x;
	if(x==y) return 1;
	if(x==fa[x]) return 0;
	return find(fa[x],cnt,y);
}

int main()
{
	int n,m; cin>>n>>m;
	int cnt = 0;
	int ans = INT_MAX;
	for(int i=0;i<=n;++i) fa[i] = i;
	for(int i=0;i<m;++i)
	{
		int u,v;
		cin>>u>>v;
		cnt = 0;
		if(find(v,cnt,u))
		{
			if(cnt<ans)
			{
				ans = cnt;
				for(int i=0;i<cnt;++i) a[i] = b[i];
			}
		}
		else fa[u] = v;
		
	}
	if(ans==INT_MAX) {
		cout<<"-1"<<"\n"; return 0;
	}
	cout<<ans<<"\n";
	for(int i=0;i<ans;++i) cout<<a[i]<<"\n";
}