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;

#define ff first
#define ss second
#define pb push_back

using ll = long long;
using ii = pair<int, int>;

const int oo = 1e9;
const int N = 1e3 + 5;

int p[N];
vector<ii> adj[N];
int r[N];

int fid(int x){
	return r[x] = x == r[x] ? x : fid(r[x]);
}

void joi(int a, int b){
	r[fid(a)] = fid(b);
}

bool bfs(int ini){
	queue<int> q;
	q.push(ini);
	p[ini] = -1;
	while(!q.empty()){
		int v = q.front();
		q.pop();
		bool ok = 1;
		for(ii i : adj[v]){
			if(i.ff != ini && p[i.ff] != 0){
				if(fid(v) == fid(i.ff)){
					ok = 0;
					break;
				}
			}
		}
		if(ok){
			for(ii i : adj[v]){
				if(i.ff == ini){
					vector<int> ans;
					while(v != -1){
						ans.pb(v);
						v = p[v];
					}
					printf("%d\n", ans.size());
					for(int j : ans){
						printf("%d\n", j);
					}
					return 1;
				}
				else if(!p[i.ff]){
					if(v != 1){
						joi(v, i.ff);
					}
					p[i.ff] = v;
					q.push(i.ff);
				}
			}
		}
	}
	return false;
}

int main(){
	int n, m;
	scanf("%d %d", &n, &m);
	for(int i = 0; i < m; i++){
		int a, b;
		scanf("%d %d", &a, &b);
		adj[a].pb({b, i});
	}
	for(int i = 1; i <= n; i++){
		memset(p, 0, sizeof(p));
		for(int j = 1; j <= n; j++){
			r[j] = j;
		}
		if(bfs(i)) return 0;
	}
	printf("-1\n");
	return 0;
}