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

const int MAX_V=300005;
vector<int> G[MAX_V];
int d[MAX_V];
int V;

void bfs(int s){
  queue<int> q;
  q.push(s);
  rep(i,3*V) d[i]=-1;
  d[s]=0;
  while(q.size()){
    int v1=q.front(); q.pop();
    for(int v2:G[v1]){
      if(d[v2]!=-1) continue;
      d[v2]=d[v1]+1;
      q.push(v2);
    }
  }
}

int main(){
  int m;
  cin>>V>>m;
  rep(i,m){
    int u,v;
    cin>>u>>v;
    u--; v--;
    rep(j,3){
      G[3*u+j].push_back(3*v+(j+1)%3);
    }
  }
  int s,t;
  cin>>s>>t;
  s--; t--;
  bfs(3*s);
  if(d[3*t]==-1) cout<<-1<<endl;
  else cout<<(d[3*t]+1)/3<<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>
#define rep(i,n) for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

const int MAX_V=300005;
vector<int> G[MAX_V];
int d[MAX_V];
int V;

void bfs(int s){
  queue<int> q;
  q.push(s);
  rep(i,3*V) d[i]=-1;
  d[s]=0;
  while(q.size()){
    int v1=q.front(); q.pop();
    for(int v2:G[v1]){
      if(d[v2]!=-1) continue;
      d[v2]=d[v1]+1;
      q.push(v2);
    }
  }
}

int main(){
  int m;
  cin>>V>>m;
  rep(i,m){
    int u,v;
    cin>>u>>v;
    u--; v--;
    rep(j,3){
      G[3*u+j].push_back(3*v+(j+1)%3);
    }
  }
  int s,t;
  cin>>s>>t;
  s--; t--;
  bfs(3*s);
  if(d[3*t]==-1) cout<<-1<<endl;
  else cout<<(d[3*t]+1)/3<<endl;
  return 0;
}




```