#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+5;

#define endl "\n"

bool vis[N][3];
vector<int> G[N];
int n,m,S,T;

struct node{
	int now,t;
};

void bfs(){
	node p,q;
	p.now=S;
	p.t=0;
	vis[p.now][p.t]=1;
	queue<node> Q;
	Q.push(p);
	while(!Q.empty()){
		q=Q.front();
		Q.pop();
		if(q.now==T&&q.t%3==0) {
			cout<<q.t/3<<endl;
			return;
		}
		int t=(q.t+1)%3,v=q.now,u;
		for(int i=0;i<G[v].size();i++){
			u=G[v][i];
			if(!vis[u][t]){
				vis[u][t]=1;
				p.now=u;
				p.t=q.t+1;
				Q.push(p);
			}
		}
	}
	cout<<-1<<endl;
}

int main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	cin>>n>>m;
	memset(vis,0,sizeof(vis));
	int v,u;
	while(m--){
		cin>>v>>u;
		G[v].push_back(u);
	}
	cin>>S>>T;
	bfs();
	return 0;
}
### Prompt

Please create a solution in cpp to the following problem:

Ken loves ken-ken-pa (Japanese version of hopscotch). Today, he will play it on a directed graph G. G consists of N vertices numbered 1 to N, and M edges. The i-th edge points from Vertex u_i to Vertex v_i.

First, Ken stands on Vertex S. He wants to reach Vertex T by repeating ken-ken-pa. In one ken-ken-pa, he does the following exactly three times: follow an edge pointing from the vertex on which he is standing.

Determine if he can reach Vertex T by repeating ken-ken-pa. If the answer is yes, find the minimum number of ken-ken-pa needed to reach Vertex T. Note that visiting Vertex T in the middle of a ken-ken-pa does not count as reaching Vertex T by repeating ken-ken-pa.

Constraints

* 2 \leq N \leq 10^5
* 0 \leq M \leq \min(10^5, N (N-1))
* 1 \leq u_i, v_i \leq N(1 \leq i \leq M)
* u_i \neq v_i (1 \leq i \leq M)
* If i \neq j, (u_i, v_i) \neq (u_j, v_j).
* 1 \leq S, T \leq N
* S \neq T

Input

Input is given from Standard Input in the following format:


N M
u_1 v_1
:
u_M v_M
S T


Output

If Ken cannot reach Vertex T from Vertex S by repeating ken-ken-pa, print -1. If he can, print the minimum number of ken-ken-pa needed to reach vertex T.

Examples

Input

4 4
1 2
2 3
3 4
4 1
1 3


Output

2


Input

3 3
1 2
2 3
3 1
1 2


Output

-1


Input

2 0
1 2


Output

-1


Input

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6


Output

2

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+5;

#define endl "\n"

bool vis[N][3];
vector<int> G[N];
int n,m,S,T;

struct node{
	int now,t;
};

void bfs(){
	node p,q;
	p.now=S;
	p.t=0;
	vis[p.now][p.t]=1;
	queue<node> Q;
	Q.push(p);
	while(!Q.empty()){
		q=Q.front();
		Q.pop();
		if(q.now==T&&q.t%3==0) {
			cout<<q.t/3<<endl;
			return;
		}
		int t=(q.t+1)%3,v=q.now,u;
		for(int i=0;i<G[v].size();i++){
			u=G[v][i];
			if(!vis[u][t]){
				vis[u][t]=1;
				p.now=u;
				p.t=q.t+1;
				Q.push(p);
			}
		}
	}
	cout<<-1<<endl;
}

int main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	cin>>n>>m;
	memset(vis,0,sizeof(vis));
	int v,u;
	while(m--){
		cin>>v>>u;
		G[v].push_back(u);
	}
	cin>>S>>T;
	bfs();
	return 0;
}
```