#include <bits/stdc++.h>
using namespace std;
using P = pair<int,int>;

int main(){
  int n, m, u, v;
  cin >> n >> m;
  vector<vector<int>> ver(n,vector<int>(0));
  for(int i=0; i<m; ++i){
    cin >> u >> v;
    --u; --v;
    ver[u].push_back(v);
  }
  int s, t;
  cin >> s >> t;
  --s; --t;
  
  vector<vector<bool>> ng(n,vector<bool>(3,false));
  queue<P> que;
  que.push(make_pair(s,0));
  ng[s][0] = true;
  while(!que.empty()){
    int x = que.front().first;
    int dis = que.front().second;
    if(x == t && (dis%3) == 0){
      cout << dis/3 << endl;
      return 0;
    }
    que.pop();
    int ndis = dis+1;
    for(int a: ver[x]){
      if(ng[a][ndis%3]) continue;
      ng[a][ndis%3] = true;
      que.push(make_pair(a,ndis));
    }
  }
  cout << -1 << 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
#include <bits/stdc++.h>
using namespace std;
using P = pair<int,int>;

int main(){
  int n, m, u, v;
  cin >> n >> m;
  vector<vector<int>> ver(n,vector<int>(0));
  for(int i=0; i<m; ++i){
    cin >> u >> v;
    --u; --v;
    ver[u].push_back(v);
  }
  int s, t;
  cin >> s >> t;
  --s; --t;
  
  vector<vector<bool>> ng(n,vector<bool>(3,false));
  queue<P> que;
  que.push(make_pair(s,0));
  ng[s][0] = true;
  while(!que.empty()){
    int x = que.front().first;
    int dis = que.front().second;
    if(x == t && (dis%3) == 0){
      cout << dis/3 << endl;
      return 0;
    }
    que.pop();
    int ndis = dis+1;
    for(int a: ver[x]){
      if(ng[a][ndis%3]) continue;
      ng[a][ndis%3] = true;
      que.push(make_pair(a,ndis));
    }
  }
  cout << -1 << endl;
}
```