#include"bits/stdc++.h"
using namespace std;
typedef long long ll;
#define int ll
vector<vector<int>> g;
int n,m,s,t;
int32_t main()
{
	#ifndef ONLINE_JUDGE
	freopen("in","r",stdin);
	freopen("out","w",stdout);
	#endif
	int u,v;
	cin>>n>>m;
	g=vector<vector<int>>(3*n+1);
	for(int i=0;i<m;i++)
	{
		cin>>u>>v;
		g[u].push_back(v+n);
		g[u+n].push_back(v+2*n);
		g[u+2*n].push_back(v);
	}
	cin>>s>>t;

	queue<int> q;
	int cur=0;q.push(s);
	vector<int> dist(3*n+1,1e9);
	vector<bool> vis(3*n+1);
	dist[s]=0;
	while(!q.empty())
	{
		int x=q.front();
		q.pop();
		for(auto u:g[x])
		{
			if(vis[u])continue;
			dist[u]=min(dist[u],dist[x]+1);
			q.push(u);vis[u]=1;
		}

	}

	if(dist[t]==1e9)cout<<-1;
	else cout<<dist[t]/3;


	

}
### Prompt

Develop a solution in cpp 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;
typedef long long ll;
#define int ll
vector<vector<int>> g;
int n,m,s,t;
int32_t main()
{
	#ifndef ONLINE_JUDGE
	freopen("in","r",stdin);
	freopen("out","w",stdout);
	#endif
	int u,v;
	cin>>n>>m;
	g=vector<vector<int>>(3*n+1);
	for(int i=0;i<m;i++)
	{
		cin>>u>>v;
		g[u].push_back(v+n);
		g[u+n].push_back(v+2*n);
		g[u+2*n].push_back(v);
	}
	cin>>s>>t;

	queue<int> q;
	int cur=0;q.push(s);
	vector<int> dist(3*n+1,1e9);
	vector<bool> vis(3*n+1);
	dist[s]=0;
	while(!q.empty())
	{
		int x=q.front();
		q.pop();
		for(auto u:g[x])
		{
			if(vis[u])continue;
			dist[u]=min(dist[u],dist[x]+1);
			q.push(u);vis[u]=1;
		}

	}

	if(dist[t]==1e9)cout<<-1;
	else cout<<dist[t]/3;


	

}
```