#include <bits/stdc++.h>
using namespace std;
int INF = 10000000;
int main(){
  int N, M;
  cin >> N >> M;
  vector<vector<int>> E(N * 3);
  for (int i = 0; i < M; i++){
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    E[u * 3].push_back(v * 3 + 1);
    E[u * 3 + 1].push_back(v * 3 + 2);
    E[u * 3 + 2].push_back(v * 3);
  }
  int S, T;
  cin >> S >> T;
  S--;
  T--;
  vector<int> d(N * 3, INF);
  d[S * 3] = 0;
  queue<int> Q;
  Q.push(S * 3);
  while (!Q.empty()){
    int v = Q.front();
    Q.pop();
    for (auto w : E[v]){
      if (d[w] == INF){
        d[w] = d[v] + 1;
        Q.push(w);
      }
    }
  }
  if (d[T * 3] == INF){
    cout << -1 << endl;
  } else {
    cout << d[T * 3] / 3 << endl;
  }
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 INF = 10000000;
int main(){
  int N, M;
  cin >> N >> M;
  vector<vector<int>> E(N * 3);
  for (int i = 0; i < M; i++){
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    E[u * 3].push_back(v * 3 + 1);
    E[u * 3 + 1].push_back(v * 3 + 2);
    E[u * 3 + 2].push_back(v * 3);
  }
  int S, T;
  cin >> S >> T;
  S--;
  T--;
  vector<int> d(N * 3, INF);
  d[S * 3] = 0;
  queue<int> Q;
  Q.push(S * 3);
  while (!Q.empty()){
    int v = Q.front();
    Q.pop();
    for (auto w : E[v]){
      if (d[w] == INF){
        d[w] = d[v] + 1;
        Q.push(w);
      }
    }
  }
  if (d[T * 3] == INF){
    cout << -1 << endl;
  } else {
    cout << d[T * 3] / 3 << endl;
  }
}
```