#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000000
#define P pair<int,int>
#define pb push_back
#define fi first
#define se second
int n,m,st,d[1010];
vector<int>v[1010],vg[1010],ans;
P es[2010];
int mid=INF,mi;
queue<int>q;
void dfs(int x){
	if(d[x]==0)return;
	for(int i:vg[x]){
		if(d[i]==d[x]-1){
			ans.pb(i);
			dfs(i);
			break;
		}
	}
}
int main(){
	cin>>n>>m;
	rep(i,m){
		int a,b;
		cin>>a>>b;
		a--,b--;
		v[a].pb(b);
		vg[b].pb(a);
		es[i]={b,a};
	}
	rep(i,m){
		P e=es[i];
		fill(d,d+n,INF);
		d[e.fi]=0;
		q.push(e.fi);
		while(!q.empty()){
			int p=q.front();
			q.pop();
			for(int j:v[p]){
				if(d[j]>d[p]+1){
					d[j]=d[p]+1;
					q.push(j);
				}
			}
		}
		if(mid>d[e.se]){
			mid=d[e.se];
			mi=i;
		}
	}
	if(mid==INF){
		puts("-1");
		return 0;
	}
	P e=es[mi];
	fill(d,d+n,INF);
	d[e.fi]=0;
	q.push(e.fi);
	while(!q.empty()){
		int p=q.front();
		q.pop();
		for(int j:v[p]){
			if(d[j]>d[p]+1){
				d[j]=d[p]+1;
				q.push(j);
			}
		}
	}
	ans.pb(e.se);
	dfs(e.se);
	cout<<ans.size()<<endl;
	for(int i:ans){
		cout<<i+1<<endl;
	}
	return 0;
}

### Prompt

Please formulate 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;
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000000
#define P pair<int,int>
#define pb push_back
#define fi first
#define se second
int n,m,st,d[1010];
vector<int>v[1010],vg[1010],ans;
P es[2010];
int mid=INF,mi;
queue<int>q;
void dfs(int x){
	if(d[x]==0)return;
	for(int i:vg[x]){
		if(d[i]==d[x]-1){
			ans.pb(i);
			dfs(i);
			break;
		}
	}
}
int main(){
	cin>>n>>m;
	rep(i,m){
		int a,b;
		cin>>a>>b;
		a--,b--;
		v[a].pb(b);
		vg[b].pb(a);
		es[i]={b,a};
	}
	rep(i,m){
		P e=es[i];
		fill(d,d+n,INF);
		d[e.fi]=0;
		q.push(e.fi);
		while(!q.empty()){
			int p=q.front();
			q.pop();
			for(int j:v[p]){
				if(d[j]>d[p]+1){
					d[j]=d[p]+1;
					q.push(j);
				}
			}
		}
		if(mid>d[e.se]){
			mid=d[e.se];
			mi=i;
		}
	}
	if(mid==INF){
		puts("-1");
		return 0;
	}
	P e=es[mi];
	fill(d,d+n,INF);
	d[e.fi]=0;
	q.push(e.fi);
	while(!q.empty()){
		int p=q.front();
		q.pop();
		for(int j:v[p]){
			if(d[j]>d[p]+1){
				d[j]=d[p]+1;
				q.push(j);
			}
		}
	}
	ans.pb(e.se);
	dfs(e.se);
	cout<<ans.size()<<endl;
	for(int i:ans){
		cout<<i+1<<endl;
	}
	return 0;
}

```