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 <cstdio>
#include<string.h>
#include <algorithm>
#include<vector>
using namespace std;
const int maxn = 5000+10;
bool g[1010][1010]= {0};
struct Edge {
	int to, nxt;
} e[maxn];
int head[maxn], cnte=1;
int n, m;
inline void add(const int& from, const int& to) {
	++cnte;
	e[cnte].to = to;
	e[cnte].nxt = head[from];
	head[from] = cnte;
}
int path[maxn];
int vi;
int depth[maxn];
void clear() {
	memset(path,0,sizeof(path));
	memset(depth,-1,sizeof(depth));
}
int sum=100000;
vector<int> ans;

void dfs(int u,int f,int dis) {
	depth[u]=dis;
	for(int i=head[u]; i; i=e[i].nxt) {
		
		int nv = e[i].to;
		if(depth[nv]==-1) {
			path[nv]=u;
			dfs(nv,f,dis+1);
		} else {
			if(nv==f) {
				path[nv]=u;
				if(sum>depth[u]) {
					ans.clear();
					sum=depth[u];
					while(path[nv]!=f) {
						ans.push_back(nv);
						nv=path[nv];
					}
					ans.push_back(nv);
				}
			} else if(depth[nv]>depth[u]+1) {
				path[nv]=u;
				dfs(nv,f,dis+1);
			}
		}
	}
}
int main() {
	memset(head,0,sizeof(head));
	cin>>n>>m;
	for(int i=0; i<m; i++) {
		int u,v;
		cin>>u>>v;
		g[u][v]=1;
		add(u,v);
	}
	for(int i=1; i<=n; i++) {
		clear();
		dfs(i,i,1);
	}
	if(sum==1e5)cout<<"-1";
	else {
		cout<<sum<<endl;
		for(int i=0; i<ans.size(); i++)
			cout<<ans[i]<<endl;
	}
	return 0;
}