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

signed main() {
  int n,m;
  cin >> n >> m;

  vector<vector<int>> G(n);
  for (int i = 0; i < m; i++) {
    int u,v;
    cin >> u >> v;
    u--, v--;

    G[u].push_back(v);
  }
  int s,t;
  cin >> s >> t;
  s--, t--;

  vector<vector<int>> dist(n, vector<int> (3,-1));
  queue<pair<int,int>> que;
  que.emplace(s,0);
  dist[s][0] = 0;
  while(!que.empty()) {
    int cur = que.front().first;
    int mod = que.front().second;
    que.pop();

    for (auto nxt : G[cur]) {
      if(dist[nxt][(mod+1)%3] != -1) continue;

      que.emplace(nxt, (mod+1)%3);
      dist[nxt][(mod+1)%3] = dist[cur][mod] + 1;
    }
  }
  int ans = dist[t][0];
  if(ans != -1) ans /= 3;
  cout << ans << '\n';
  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;

signed main() {
  int n,m;
  cin >> n >> m;

  vector<vector<int>> G(n);
  for (int i = 0; i < m; i++) {
    int u,v;
    cin >> u >> v;
    u--, v--;

    G[u].push_back(v);
  }
  int s,t;
  cin >> s >> t;
  s--, t--;

  vector<vector<int>> dist(n, vector<int> (3,-1));
  queue<pair<int,int>> que;
  que.emplace(s,0);
  dist[s][0] = 0;
  while(!que.empty()) {
    int cur = que.front().first;
    int mod = que.front().second;
    que.pop();

    for (auto nxt : G[cur]) {
      if(dist[nxt][(mod+1)%3] != -1) continue;

      que.emplace(nxt, (mod+1)%3);
      dist[nxt][(mod+1)%3] = dist[cur][mod] + 1;
    }
  }
  int ans = dist[t][0];
  if(ans != -1) ans /= 3;
  cout << ans << '\n';
  return 0;
}
```