#include<bits/stdc++.h>
using namespace std;
vector<int> G[300100];
int dist[300100];
int main()
{
	int N,M;
	cin>>N>>M;
	for(int i=0;i<N*3;i++)
		dist[i]=-1;
	while(M--)
	{
		int x,y;
		cin>>x>>y;
		x--;
		y--;
		for(int i=0;i<3;i++)
			G[x*3+i].push_back(y*3+(i+1)%3);
	}
	int S,T;
	cin>>S>>T;
	S--;
	T--;
	queue<int> q;
	S*=3;
	T*=3;
	q.push(S);
	dist[S]=0;
	while(!q.empty())
	{
		int x=q.front();
		q.pop();
		for(int i=0;i<G[x].size();i++)
			if(dist[G[x][i]]==-1)
			{
				dist[G[x][i]]=dist[x]+1;
				q.push(G[x][i]);
			}
	}
	if(dist[T]==-1)
		puts("-1");
	else	cout<<dist[T]/3<<endl;
	return 0;
}
### Prompt

Generate a CPP solution 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;
vector<int> G[300100];
int dist[300100];
int main()
{
	int N,M;
	cin>>N>>M;
	for(int i=0;i<N*3;i++)
		dist[i]=-1;
	while(M--)
	{
		int x,y;
		cin>>x>>y;
		x--;
		y--;
		for(int i=0;i<3;i++)
			G[x*3+i].push_back(y*3+(i+1)%3);
	}
	int S,T;
	cin>>S>>T;
	S--;
	T--;
	queue<int> q;
	S*=3;
	T*=3;
	q.push(S);
	dist[S]=0;
	while(!q.empty())
	{
		int x=q.front();
		q.pop();
		for(int i=0;i<G[x].size();i++)
			if(dist[G[x][i]]==-1)
			{
				dist[G[x][i]]=dist[x]+1;
				q.push(G[x][i]);
			}
	}
	if(dist[T]==-1)
		puts("-1");
	else	cout<<dist[T]/3<<endl;
	return 0;
}
```