#include <bits/stdc++.h>
using namespace std;

int main() {
  int n, m, s, t;
  cin >> n >> m;
  vector<vector<int>> v(n);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    v[a - 1].push_back(b - 1);
  }
  cin >> s >> t;
  s--;
  t--;

  vector<vector<int>> d(n, vector<int>(3, 1e9));
  queue<pair<int, int>> q;
  d[s][0] = 0;
  q.push({s, 0});
  while (!q.empty()) {
    auto e = q.front();
    int i = e.first, di = e.second;
    q.pop();
    for (auto j : v[i]) {
      if (di + 1 < d[j][(di + 1) % 3]) {
        d[j][(di + 1) % 3] = di + 1;
        q.push({j, di + 1});
      }
    }
  }

  cout << (d[t][0] != int(1e9) ? d[t][0] / 3 : -1) << endl;
}

### Prompt

Your challenge is to write 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() {
  int n, m, s, t;
  cin >> n >> m;
  vector<vector<int>> v(n);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    v[a - 1].push_back(b - 1);
  }
  cin >> s >> t;
  s--;
  t--;

  vector<vector<int>> d(n, vector<int>(3, 1e9));
  queue<pair<int, int>> q;
  d[s][0] = 0;
  q.push({s, 0});
  while (!q.empty()) {
    auto e = q.front();
    int i = e.first, di = e.second;
    q.pop();
    for (auto j : v[i]) {
      if (di + 1 < d[j][(di + 1) % 3]) {
        d[j][(di + 1) % 3] = di + 1;
        q.push({j, di + 1});
      }
    }
  }

  cout << (d[t][0] != int(1e9) ? d[t][0] / 3 : -1) << endl;
}

```