#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

const int MOD=1000000007;
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;

vector<int>E[2000];
int d[2000][2000];
int prv[2000][2000];

void bfs(int v){
	d[v][v]=0;
	prv[v][v]=-1;
	queue<int>que;
	que.push(v);
	while(!que.empty()){
		int p=que.front();que.pop();
		for(int u:E[p]){
			if(d[v][u]==-1){
				d[v][u]=d[v][p]+1;
				prv[v][u]=p;
				que.push(u);
			}
		}
	}
}
int main(){
	int n,m;cin>>n>>m;
	rep(i,m){
		int a,b;scanf("%d%d",&a,&b);a--;b--;
		E[a].push_back(b);
	}
	memset(d,-1,sizeof(d));
	rep(i,n)bfs(i);
	int idx=-1,idy=-1;
	int Min=INF;
	rep(i,n)rep(j,n){
		if(i==j)continue;
		if(d[i][j]!=-1&&d[j][i]!=-1&&Min>d[i][j]+d[j][i]){
			Min=d[i][j]+d[j][i];
			idx=i;
			idy=j;
		}
	}
	if(Min==INF){
		puts("-1");return 0;
	}
	int x=idy;
	vector<int>v{idy};
	while(x!=idx){
		x=prv[idx][x];
		v.push_back(x);
	}
	while(x!=idy){
		x=prv[idy][x];
		if(x!=idy)v.push_back(x);
	}
	cout<<v.size()<<endl;
	reverse(v.begin(),v.end());
	for(int i:v){
		printf("%d\n",i+1);
	}
}
### Prompt

Create a solution in Cpp for 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>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

const int MOD=1000000007;
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;

vector<int>E[2000];
int d[2000][2000];
int prv[2000][2000];

void bfs(int v){
	d[v][v]=0;
	prv[v][v]=-1;
	queue<int>que;
	que.push(v);
	while(!que.empty()){
		int p=que.front();que.pop();
		for(int u:E[p]){
			if(d[v][u]==-1){
				d[v][u]=d[v][p]+1;
				prv[v][u]=p;
				que.push(u);
			}
		}
	}
}
int main(){
	int n,m;cin>>n>>m;
	rep(i,m){
		int a,b;scanf("%d%d",&a,&b);a--;b--;
		E[a].push_back(b);
	}
	memset(d,-1,sizeof(d));
	rep(i,n)bfs(i);
	int idx=-1,idy=-1;
	int Min=INF;
	rep(i,n)rep(j,n){
		if(i==j)continue;
		if(d[i][j]!=-1&&d[j][i]!=-1&&Min>d[i][j]+d[j][i]){
			Min=d[i][j]+d[j][i];
			idx=i;
			idy=j;
		}
	}
	if(Min==INF){
		puts("-1");return 0;
	}
	int x=idy;
	vector<int>v{idy};
	while(x!=idx){
		x=prv[idx][x];
		v.push_back(x);
	}
	while(x!=idy){
		x=prv[idy][x];
		if(x!=idy)v.push_back(x);
	}
	cout<<v.size()<<endl;
	reverse(v.begin(),v.end());
	for(int i:v){
		printf("%d\n",i+1);
	}
}
```