#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100010;
const int64_t INF = (1LL << 62);
vector<int64_t> to[MAXN];
vector<vector<int64_t>> steps(3, vector<int64_t>(MAXN, INF));

int main()
{
  int64_t N, M, u, v, S, T;
  cin >> N >> M;
  for (int i = 0; i < M; i++)
  {
    cin >> u >> v;
    u--, v--;
    to[u].push_back(v);
  }
  cin >> S >> T;
  S--, T--;
  queue<pair<int64_t, int64_t>> Q;
  Q.push({S, 0});
  while (Q.size())
  {
    int64_t cur = Q.front().first, step = Q.front().second;
    Q.pop();
    for (auto v : to[cur])
    {
      if (steps[(step + 1) % 3][v] != INF)
        continue;
      steps[(step + 1) % 3][v] = step + 1;
      Q.push({v, step + 1});
    }
  }
  cout << (steps[0][T] == INF ? -1 : steps[0][T] / 3) << 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;
const int MAXN = 100010;
const int64_t INF = (1LL << 62);
vector<int64_t> to[MAXN];
vector<vector<int64_t>> steps(3, vector<int64_t>(MAXN, INF));

int main()
{
  int64_t N, M, u, v, S, T;
  cin >> N >> M;
  for (int i = 0; i < M; i++)
  {
    cin >> u >> v;
    u--, v--;
    to[u].push_back(v);
  }
  cin >> S >> T;
  S--, T--;
  queue<pair<int64_t, int64_t>> Q;
  Q.push({S, 0});
  while (Q.size())
  {
    int64_t cur = Q.front().first, step = Q.front().second;
    Q.pop();
    for (auto v : to[cur])
    {
      if (steps[(step + 1) % 3][v] != INF)
        continue;
      steps[(step + 1) % 3][v] = step + 1;
      Q.push({v, step + 1});
    }
  }
  cout << (steps[0][T] == INF ? -1 : steps[0][T] / 3) << endl;
}
```