#include<bits/stdc++.h>
using namespace std;
int main(){
  long N,M;cin>>N>>M;
  vector<vector<long>>A(N*3);
  for(int i=0;i<M;i++){
    long u,v;cin>>u>>v;
    A[u-1].push_back(v-1+N);
    //A[v-1].push_back(u-1+N);
    A[u-1+N].push_back(v-1+2*N);
    //A[v-1+N].push_back(u-1+2*N);
    A[u-1+2*N].push_back(v-1);
    //A[v-1+2*N].push_back(u-1);
  }
  long S,T;cin>>S>>T;
  vector<long>D(3*N,-3);
  queue<long>q;q.push(S-1);D[S-1]=0;
  while(q.size()){
    long f=q.front();q.pop();
    for(long n:A[f]){
      if(D[n]==-3){D[n]=D[f]+1;q.push(n);}
      else D[n]=min(D[n],D[f]+1);
    }
  }
  //for(int i=0;i<3;i++){for(int j=0;j<N;j++)cout<<D[j+i*N]<<" ";cout<<endl;}
  //cout<<D[T-1]<<endl;
  cout<<(D[T-1]%3==0?D[T-1]/3:-1);
  
}
### Prompt

Your task is to create 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(){
  long N,M;cin>>N>>M;
  vector<vector<long>>A(N*3);
  for(int i=0;i<M;i++){
    long u,v;cin>>u>>v;
    A[u-1].push_back(v-1+N);
    //A[v-1].push_back(u-1+N);
    A[u-1+N].push_back(v-1+2*N);
    //A[v-1+N].push_back(u-1+2*N);
    A[u-1+2*N].push_back(v-1);
    //A[v-1+2*N].push_back(u-1);
  }
  long S,T;cin>>S>>T;
  vector<long>D(3*N,-3);
  queue<long>q;q.push(S-1);D[S-1]=0;
  while(q.size()){
    long f=q.front();q.pop();
    for(long n:A[f]){
      if(D[n]==-3){D[n]=D[f]+1;q.push(n);}
      else D[n]=min(D[n],D[f]+1);
    }
  }
  //for(int i=0;i<3;i++){for(int j=0;j<N;j++)cout<<D[j+i*N]<<" ";cout<<endl;}
  //cout<<D[T-1]<<endl;
  cout<<(D[T-1]%3==0?D[T-1]/3:-1);
  
}
```