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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#include <utility>
#include <functional>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
int n, m;
vector<int> g[1010], gr[1010];
bool used[1010];
int cmp[1010];
vector<int> vs;
void dfs(int x){
	used[x]=1;
	for(auto y:g[x]){
		if(!used[y]) dfs(y);
	}
	vs.push_back(x);
}
void rdfs(int v, int k){
	used[v]=1;
	cmp[v]=k;
	for(auto y:gr[v]){
		if(!used[y]) rdfs(y, k);
	}
}
int scc(){
	fill(used, used+n, 0);
	vs.clear();
	for(int i=0; i<n; i++){
		if(!used[i]) dfs(i);
	}
	fill(used, used+n, 0);
	int k=0;
	for(int i=vs.size()-1; i>=0; i--){
		if(!used[vs[i]]) rdfs(vs[i], k++);
	}
	return k;
}
int main()
{
	cin>>n>>m;
	for(int i=0; i<m; i++){
		int a, b; cin>>a>>b; a--; b--;
		g[a].push_back(b);
		gr[b].push_back(a);
	}
	int k=scc();
	if(k==n) cout<<-1<<endl;
	else cout<<0<<endl;
	return 0;
}