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

int n,m;
vector<vector<int>> e;
vector<int> length;
queue<int> vertex;

void bfs(int i){
  for(int x:e[i]){
    if(length[(x+n)%(3*n)]>length[i]+1){
      length[(x+n)%(3*n)]=length[i]+1;
      vertex.push((x+n)%(3*n));
    }
  }
}

int main(){
  cin>>n>>m;
  e.resize(3*n);
  length.resize(3*n);
  for(int i=0; i<3*n; ++i)length[i]=1000000000;
  
  for(int i=0; i<m; ++i){
    int u,v; cin>>u>>v;
    e[u-1].push_back(v-1+n);
    e[u-1+n].push_back(v-1+2*n);
    e[u-1+2*n].push_back(v-1);
  }
  int s,t; cin>>s>>t;

  length[s-1]=0;
  vertex.push(s-1);
  while(!vertex.empty()){
    bfs(vertex.front());
    vertex.pop();
  }

  if(length[t-1]==1000000000)cout<<-1<<endl;
  else cout<<length[t-1]/3<<endl;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 n,m;
vector<vector<int>> e;
vector<int> length;
queue<int> vertex;

void bfs(int i){
  for(int x:e[i]){
    if(length[(x+n)%(3*n)]>length[i]+1){
      length[(x+n)%(3*n)]=length[i]+1;
      vertex.push((x+n)%(3*n));
    }
  }
}

int main(){
  cin>>n>>m;
  e.resize(3*n);
  length.resize(3*n);
  for(int i=0; i<3*n; ++i)length[i]=1000000000;
  
  for(int i=0; i<m; ++i){
    int u,v; cin>>u>>v;
    e[u-1].push_back(v-1+n);
    e[u-1+n].push_back(v-1+2*n);
    e[u-1+2*n].push_back(v-1);
  }
  int s,t; cin>>s>>t;

  length[s-1]=0;
  vertex.push(s-1);
  while(!vertex.empty()){
    bfs(vertex.front());
    vertex.pop();
  }

  if(length[t-1]==1000000000)cout<<-1<<endl;
  else cout<<length[t-1]/3<<endl;
}

```