#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i=0;i<n;i++)
#define rep2(i,a,b) for(int i=(a);i<(b);++i)

int main() {
  int N,M,u,v,S,T;
  cin >> N >> M;
  vector<vector<int>> g(N);
  rep(i,M) {
    cin >> u >> v;
    u--;v--;
    g[u].push_back(v);
  }
  cin >> S >> T;
  S--;T--;

  priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>> pq;
  vector<vector<ll>> ans(N,vector<ll> (3,LONG_LONG_MAX/100));
  pq.push({0,S,0});//距離、場所、状態,from

  while (!pq.empty()) {
    auto t = pq.top();
    pq.pop();
    if (ans[t[1]][t[2]]>t[0]){
      ans[t[1]][t[2]] = t[0];
      for (auto e:g[t[1]]) {
        pq.push({t[0] + 1 ,e ,(t[2]+1)%3});
      }
    }
  }
  int res = LONG_LONG_MAX/100>ans[T][0]?ans[T][0]/3:-1;
  cout << res << 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;
typedef long long ll;
#define rep(i,n) for(int i=0;i<n;i++)
#define rep2(i,a,b) for(int i=(a);i<(b);++i)

int main() {
  int N,M,u,v,S,T;
  cin >> N >> M;
  vector<vector<int>> g(N);
  rep(i,M) {
    cin >> u >> v;
    u--;v--;
    g[u].push_back(v);
  }
  cin >> S >> T;
  S--;T--;

  priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>> pq;
  vector<vector<ll>> ans(N,vector<ll> (3,LONG_LONG_MAX/100));
  pq.push({0,S,0});//距離、場所、状態,from

  while (!pq.empty()) {
    auto t = pq.top();
    pq.pop();
    if (ans[t[1]][t[2]]>t[0]){
      ans[t[1]][t[2]] = t[0];
      for (auto e:g[t[1]]) {
        pq.push({t[0] + 1 ,e ,(t[2]+1)%3});
      }
    }
  }
  int res = LONG_LONG_MAX/100>ans[T][0]?ans[T][0]/3:-1;
  cout << res << endl;
}
```