#include<bits/stdc++.h>
using namespace std;
vector<int> edge[300050];
int N = 100010;
int f,t;
int main(){
  int n,m;
  cin>>n>>m;
  for(int i=0;i<m;++i){
    cin>>f>>t;
    edge[f].push_back( t + N );
    edge[f + N].push_back( t + 2*N);
    edge[f + 2*N].push_back(t);
  }
  cin>>f>>t;
  int step = 0;
  set<int> past;
  past.insert(f);
  
  set<int> last;
  last.insert(f);
  
  int result = -1;
  while(last.size()>0){
    if(last.find(t)!=last.end()){
      result = step/3;
      break;
    }
    set<int> curr;
    for(auto i:last){
      for(auto j:edge[i]){
        if(past.find(j)==past.end()){
          curr.insert(j);
          past.insert(j);
        }
      }
    }
    
    step++;
    last = curr;
  }
  cout<<result<<endl;
  return 0;
}

### 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
#include<bits/stdc++.h>
using namespace std;
vector<int> edge[300050];
int N = 100010;
int f,t;
int main(){
  int n,m;
  cin>>n>>m;
  for(int i=0;i<m;++i){
    cin>>f>>t;
    edge[f].push_back( t + N );
    edge[f + N].push_back( t + 2*N);
    edge[f + 2*N].push_back(t);
  }
  cin>>f>>t;
  int step = 0;
  set<int> past;
  past.insert(f);
  
  set<int> last;
  last.insert(f);
  
  int result = -1;
  while(last.size()>0){
    if(last.find(t)!=last.end()){
      result = step/3;
      break;
    }
    set<int> curr;
    for(auto i:last){
      for(auto j:edge[i]){
        if(past.find(j)==past.end()){
          curr.insert(j);
          past.insert(j);
        }
      }
    }
    
    step++;
    last = curr;
  }
  cout<<result<<endl;
  return 0;
}

```