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>
#pragma GCC optimize("O3")
using namespace std;
#ifdef _debug
#define out(i) cout << #i << ' ' << i
#else 
#define out(i)
#endif
using ll = long long;
using ull = unsigned long long;
using ul = unsigned;
using db = double;
const int maxn = 3000;
int n, m;
vector<int> edge[maxn];
vector<int> res;
int pos[maxn], ted[maxn];
int start;
vector<pair<int,int>> all;
bool dfs(int now){
	if(pos[now] != -1){
		vector<int> p;
		for(int i = pos[now];i < res.size();++i)p.push_back(res[i]);
		int cnt = 0;
		for(auto &k : all)if(pos[k.first] >= pos[now] and pos[k.second] >= pos[now]){
			++cnt;
		}
		if(cnt != p.size())return false;
		cout << p.size() << '\n';
		for(auto &u : p)cout << u << '\n';
		return true;
	}
	pos[now] = res.size();
	res.push_back(now);
	for(auto & i : edge[now])if(dfs(i))return true;
	pos[now] = -1;
	res.pop_back();
	return false;
}
int main(){
	ios_base::sync_with_stdio(0), cin.tie(0);
	fill(pos, pos + maxn, -1);
	cin >> n >> m;
	for(int i = 0, a, b;i < m;++i){
		cin >> a >> b;
		all.emplace_back(a, b);
		edge[a].emplace_back(b);
	}
	for(int i = 1;i <= n;++i)if(dfs(i))return 0;;
	cout << -1 << '\n';
}