//PDF解説に倣って(DFSを諦めてBFS)
#include<bits/stdc++.h>
using namespace std;
int INF=1e9;

int main(){
  int n,m,u,v,s,t;
  cin>> n >> m;
  unordered_set<int> set;
  vector<unordered_set<int>> G(3*n);
  vector<int> D(3*n,INF);
  for (int i=0; i<m; i++){
    cin>> u >> v; u--; v--;
    G[u].insert(v+n);
    G[u+n].insert(v+2*n);
    G[u+2*n].insert(v); 
  }
  cin>> s >> t; s--; t--;
  
  D[s]=0;
  queue<int> Q; Q.push(s);
  while (!Q.empty()){
    u=Q.front(); Q.pop();
    for (auto i: G[u]){
      if (D[i]==INF){
        D[i]=D[u]+1;
        Q.push(i);
      }
    }
    if (u==t) break;
  }
  
  int ans= D[t]==INF ? -1:D[t]/3;
  cout<< ans <<endl;
}
### Prompt

Please create a solution in CPP 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
//PDF解説に倣って(DFSを諦めてBFS)
#include<bits/stdc++.h>
using namespace std;
int INF=1e9;

int main(){
  int n,m,u,v,s,t;
  cin>> n >> m;
  unordered_set<int> set;
  vector<unordered_set<int>> G(3*n);
  vector<int> D(3*n,INF);
  for (int i=0; i<m; i++){
    cin>> u >> v; u--; v--;
    G[u].insert(v+n);
    G[u+n].insert(v+2*n);
    G[u+2*n].insert(v); 
  }
  cin>> s >> t; s--; t--;
  
  D[s]=0;
  queue<int> Q; Q.push(s);
  while (!Q.empty()){
    u=Q.front(); Q.pop();
    for (auto i: G[u]){
      if (D[i]==INF){
        D[i]=D[u]+1;
        Q.push(i);
      }
    }
    if (u==t) break;
  }
  
  int ans= D[t]==INF ? -1:D[t]/3;
  cout<< ans <<endl;
}
```