#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;
//const ll MOD=998244353;

vector<vector<int>> G(101010);
ll dist[101010][3];
int N,M,S,T;

int main(){
  cin>>N>>M;
  rep(i,M){
    int u,v; cin>>u>>v; u--,v--;
    G[u].push_back(v);
  }
  cin>>S>>T; S--,T--;
  
  rep(i,N) rep(j,3) dist[i][j]=-1;
  queue<pair<int,int>> que;
  que.push({S,0});
  dist[S][0]=0;
  while(que.size()){
    auto p=que.front(); que.pop();
    int v=p.first,cur=p.second,nex=(cur+1)%3;
    if(v==T&&cur==0){
      cout<<dist[T][0]/3<<endl;
      return 0;
    }
    for(int u:G[v]){
      if(dist[u][nex]==-1){
        dist[u][nex]=dist[v][cur]+1;
        que.push({u,nex});
      }
    }
  }
  cout<<-1<<endl;
  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;
typedef long long ll;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;
//const ll MOD=998244353;

vector<vector<int>> G(101010);
ll dist[101010][3];
int N,M,S,T;

int main(){
  cin>>N>>M;
  rep(i,M){
    int u,v; cin>>u>>v; u--,v--;
    G[u].push_back(v);
  }
  cin>>S>>T; S--,T--;
  
  rep(i,N) rep(j,3) dist[i][j]=-1;
  queue<pair<int,int>> que;
  que.push({S,0});
  dist[S][0]=0;
  while(que.size()){
    auto p=que.front(); que.pop();
    int v=p.first,cur=p.second,nex=(cur+1)%3;
    if(v==T&&cur==0){
      cout<<dist[T][0]/3<<endl;
      return 0;
    }
    for(int u:G[v]){
      if(dist[u][nex]==-1){
        dist[u][nex]=dist[v][cur]+1;
        que.push({u,nex});
      }
    }
  }
  cout<<-1<<endl;
  return 0;
}
```