#include<iostream>
#include<queue>
using namespace std;

int n,m,S,T;
vector<vector<int>>graph(333333);
vector<int>dist(333333,-1);

void BFS()
{
	queue<int> que;
	dist[3*S]=0;
	que.push(3*S);
	while(!que.empty())
	{
		int f=que.front();
		que.pop();
		for(auto t:graph[f])
		{
			if(dist[t]==-1)
			{
				dist[t]=dist[f]+1;
				que.push(t);
			}
		}
	}
}

int main()
{
	cin>>n>>m;
	for(int i=0;i<m;i++)
	{
		int u,v;
		cin>>u>>v;
		u--;
		v--;
		for(int k=0;k<3;k++)
			graph[3*u+k].push_back(3*v+((k+1)%3));
	}
	cin>>S>>T;
	S--;
	T--;
	BFS();
	if(dist[3*T]==-1)cout << "-1" << endl;
	else cout << dist[3*T]/3 << endl;
	return 0;
}

### Prompt

Your challenge is to write 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<iostream>
#include<queue>
using namespace std;

int n,m,S,T;
vector<vector<int>>graph(333333);
vector<int>dist(333333,-1);

void BFS()
{
	queue<int> que;
	dist[3*S]=0;
	que.push(3*S);
	while(!que.empty())
	{
		int f=que.front();
		que.pop();
		for(auto t:graph[f])
		{
			if(dist[t]==-1)
			{
				dist[t]=dist[f]+1;
				que.push(t);
			}
		}
	}
}

int main()
{
	cin>>n>>m;
	for(int i=0;i<m;i++)
	{
		int u,v;
		cin>>u>>v;
		u--;
		v--;
		for(int k=0;k<3;k++)
			graph[3*u+k].push_back(3*v+((k+1)%3));
	}
	cin>>S>>T;
	S--;
	T--;
	BFS();
	if(dist[3*T]==-1)cout << "-1" << endl;
	else cout << dist[3*T]/3 << endl;
	return 0;
}

```