#include <bits/stdc++.h>
using namespace std;

int main(int argc, char const *argv[])
{
	int n,m;
	cin>>n>>m;
	set<int> edge[n];
	int u,v;
	for (int i = 0; i < m; ++i)
	{
		cin>>u>>v;--u;--v;
		edge[u].insert(v);
	}
	int s,t;
	cin>>s>>t;--s;--t;
	int memo[n][3];
	for (int i = 0; i < n; ++i)
	{
		memo[i][0]=-1;
		memo[i][1]=-1;
		memo[i][2]=-1;
	}
	memo[s][0]=0;
	queue<int> que;
	que.push(3*s);
	while(que.size()>0)
	{
		int now=que.front();
		que.pop();
		for (auto i = edge[now/3].begin(); i != edge[now/3].end(); ++i)
		{
			if (memo[*i][(now+1)%3]==-1)
			{
				memo[*i][(now+1)%3]=memo[now/3][now%3]+1;
				que.push((*i)*3+((now+1)%3));
			}
		}
	}
	if (memo[t][0]>0)
	{
		memo[t][0]/=3;
	}
	cout<<memo[t][0]<<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 <bits/stdc++.h>
using namespace std;

int main(int argc, char const *argv[])
{
	int n,m;
	cin>>n>>m;
	set<int> edge[n];
	int u,v;
	for (int i = 0; i < m; ++i)
	{
		cin>>u>>v;--u;--v;
		edge[u].insert(v);
	}
	int s,t;
	cin>>s>>t;--s;--t;
	int memo[n][3];
	for (int i = 0; i < n; ++i)
	{
		memo[i][0]=-1;
		memo[i][1]=-1;
		memo[i][2]=-1;
	}
	memo[s][0]=0;
	queue<int> que;
	que.push(3*s);
	while(que.size()>0)
	{
		int now=que.front();
		que.pop();
		for (auto i = edge[now/3].begin(); i != edge[now/3].end(); ++i)
		{
			if (memo[*i][(now+1)%3]==-1)
			{
				memo[*i][(now+1)%3]=memo[now/3][now%3]+1;
				que.push((*i)*3+((now+1)%3));
			}
		}
	}
	if (memo[t][0]>0)
	{
		memo[t][0]/=3;
	}
	cout<<memo[t][0]<<endl;
	return 0;
}

```