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

vector<int> G[100000];
int n,m;
ll dis[100000][3];


int main()
{
	cin >> n >> m;
	for(int i = 0;i<m;i++)
	{
		int a,b;
		cin >> a >> b;
		a--;b--;
		G[a].push_back(b);
	}
	int s,t;
	cin >> s >> t;
	s--;t--;
	for(int i = 0;i<100000;i++)
	{
		for(int j = 0;j<3;j++)dis[i][j] = -1;
	}
	using P = pair<ll,ll>;
	queue<P> Q;
	Q.emplace(s,0);
	dis[s][0] = 0;
	while(Q.size())
	{
		auto now = Q.front();
		int nken = (now.second+1)%3;
		Q.pop();
		for(auto &i:G[now.first])
		{
			if(dis[i][nken]==-1)
			{
				if(now.second==0)dis[i][nken] = dis[now.first][now.second] +1;
				else dis[i][nken] = dis[now.first][now.second];
				Q.emplace(i,nken);
			}
		}
	}
	cout << dis[t][0] << endl;

}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
using ll = long long;

vector<int> G[100000];
int n,m;
ll dis[100000][3];


int main()
{
	cin >> n >> m;
	for(int i = 0;i<m;i++)
	{
		int a,b;
		cin >> a >> b;
		a--;b--;
		G[a].push_back(b);
	}
	int s,t;
	cin >> s >> t;
	s--;t--;
	for(int i = 0;i<100000;i++)
	{
		for(int j = 0;j<3;j++)dis[i][j] = -1;
	}
	using P = pair<ll,ll>;
	queue<P> Q;
	Q.emplace(s,0);
	dis[s][0] = 0;
	while(Q.size())
	{
		auto now = Q.front();
		int nken = (now.second+1)%3;
		Q.pop();
		for(auto &i:G[now.first])
		{
			if(dis[i][nken]==-1)
			{
				if(now.second==0)dis[i][nken] = dis[now.first][now.second] +1;
				else dis[i][nken] = dis[now.first][now.second];
				Q.emplace(i,nken);
			}
		}
	}
	cout << dis[t][0] << endl;

}

```