#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <queue> 
using namespace std;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n,m;
	cin>>n>>m;
	int i,j,u,v;
	vector<vector<int> > adj;
	adj.resize(3*n+1);
	for(i=0;i<m;i++)
	{
		cin>>u>>v;
		adj[u].push_back(n+v);
		adj[n+u].push_back(2*n+v);
		adj[2*n+u].push_back(v);
	}
	int s,t;
	cin>>s>>t;
	vector<int> visit;
	visit.resize(3*n+1);
	for(i=1;i<=(3*n);i++)
		visit[i] = -1;
	queue<int> q;
	q.push(s);
	visit[s] = 0;
	while(!q.empty())
	{
		int cur = q.front();
		q.pop();
		int d = visit[cur];
		for(i=0;i<adj[cur].size();i++)
		{
			int x = adj[cur][i];
			if(visit[x]==-1)
			{
				visit[x] = d+1;
				q.push(x);
			}
		}
	}
	if(visit[t]==-1)
		cout<<"-1\n";
	else
		cout<<visit[t]/3<<'\n';
	return 0;
}
### Prompt

Please formulate 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 <vector>
#include <string>
#include <algorithm>
#include <queue> 
using namespace std;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n,m;
	cin>>n>>m;
	int i,j,u,v;
	vector<vector<int> > adj;
	adj.resize(3*n+1);
	for(i=0;i<m;i++)
	{
		cin>>u>>v;
		adj[u].push_back(n+v);
		adj[n+u].push_back(2*n+v);
		adj[2*n+u].push_back(v);
	}
	int s,t;
	cin>>s>>t;
	vector<int> visit;
	visit.resize(3*n+1);
	for(i=1;i<=(3*n);i++)
		visit[i] = -1;
	queue<int> q;
	q.push(s);
	visit[s] = 0;
	while(!q.empty())
	{
		int cur = q.front();
		q.pop();
		int d = visit[cur];
		for(i=0;i<adj[cur].size();i++)
		{
			int x = adj[cur][i];
			if(visit[x]==-1)
			{
				visit[x] = d+1;
				q.push(x);
			}
		}
	}
	if(visit[t]==-1)
		cout<<"-1\n";
	else
		cout<<visit[t]/3<<'\n';
	return 0;
}
```