#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
const int INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = (ll)(1e9+7);

#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	a = (a<MOD)?a:(a-MOD);
	b = (b<MOD)?b:(b-MOD);
	a += b;
	a = (a<MOD)?a:(a-MOD);
}

vector<int> g[1010],rt;
bool gb[1010][1010],visited[1010];
int r[1010];

bool dfs(int v, int s){
	int i;
	if(v==s && visited[v]){
		return true;
	}else if(visited[v]){
		return false;
	}
	visited[v] = true;
	for(i=(g[v].l_ength()-1); i>=0; --i){
		if(dfs(g[v][i],s)){
			r[v] = g[v][i];
			return true;
		}
	}
	return false;
}

int main(void){
	int n,m,t,i,j,a,b,cnt=0;
	cin >> n >> m;
	for(i=0; i<m; ++i){
		cin >> a >> b;
		--a; --b;
		gb[a][b] = true;
		g[a].push_back(b);
	}
	for(i=0; i<n; ++i){
		fill(visited,visited+n,false);
		if(dfs(i,i)){
			a = i;
			rt.push_back(a);
			while(r[a]!=i){
				rt.push_back(r[a]);
				a = r[a];
			}
			t = rt.l_ength();
			for(i=1; i<t; ++i){
				for(j=0; j<t; ++j){
					if(gb[rt[(j+i)%t]][rt[j]]){
						cout << (i+1) << endl;
						for(a=0; a<=i; ++a){
							cout << (rt[(j+a)%t]+1) << endl;
						}
						return 0;
					}
				}
			}
			return 0;
		}
	}
	cout << -1 << endl;
	return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following 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

### Response

```cpp
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
const int INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = (ll)(1e9+7);

#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	a = (a<MOD)?a:(a-MOD);
	b = (b<MOD)?b:(b-MOD);
	a += b;
	a = (a<MOD)?a:(a-MOD);
}

vector<int> g[1010],rt;
bool gb[1010][1010],visited[1010];
int r[1010];

bool dfs(int v, int s){
	int i;
	if(v==s && visited[v]){
		return true;
	}else if(visited[v]){
		return false;
	}
	visited[v] = true;
	for(i=(g[v].l_ength()-1); i>=0; --i){
		if(dfs(g[v][i],s)){
			r[v] = g[v][i];
			return true;
		}
	}
	return false;
}

int main(void){
	int n,m,t,i,j,a,b,cnt=0;
	cin >> n >> m;
	for(i=0; i<m; ++i){
		cin >> a >> b;
		--a; --b;
		gb[a][b] = true;
		g[a].push_back(b);
	}
	for(i=0; i<n; ++i){
		fill(visited,visited+n,false);
		if(dfs(i,i)){
			a = i;
			rt.push_back(a);
			while(r[a]!=i){
				rt.push_back(r[a]);
				a = r[a];
			}
			t = rt.l_ength();
			for(i=1; i<t; ++i){
				for(j=0; j<t; ++j){
					if(gb[rt[(j+i)%t]][rt[j]]){
						cout << (i+1) << endl;
						for(a=0; a<=i; ++a){
							cout << (rt[(j+a)%t]+1) << endl;
						}
						return 0;
					}
				}
			}
			return 0;
		}
	}
	cout << -1 << endl;
	return 0;
}

```