#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define ll long long
int vis[300008];
ll dist[300008];
vector<int> e[300008];
int n;
void add(int u, int v)
{
  e[u].pb(v+n);
  e[u+n].pb(v+2*n);
  e[u+2*n].pb(v);
}

int main()
{
  int m,i,j;
  cin>>n>>m;
  for(i=0;i<m;i++)
  {
    int u,v;
    cin>>u>>v;
    add(u,v);
  }
  int s,t;
  cin>>s>>t;
  vis[s]=1;
  queue<int> q;
  q.push(s);
  int f=0;
  while(!q.empty())
  {
    int u = q.front();
    q.pop();
    if(u==t) break;
    for(auto v: e[u])
    {
      if(!vis[v])
      {
		q.push(v);
		vis[v]=1;
        dist[v]=dist[u]+1;
      }
    }

  }
  if(dist[t]!=0 && dist[t]%3==0) {
  cout<<dist[t]/3;
  }
  else cout<<-1;
  return 0;
}
### 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;
#define pb push_back
#define ll long long
int vis[300008];
ll dist[300008];
vector<int> e[300008];
int n;
void add(int u, int v)
{
  e[u].pb(v+n);
  e[u+n].pb(v+2*n);
  e[u+2*n].pb(v);
}

int main()
{
  int m,i,j;
  cin>>n>>m;
  for(i=0;i<m;i++)
  {
    int u,v;
    cin>>u>>v;
    add(u,v);
  }
  int s,t;
  cin>>s>>t;
  vis[s]=1;
  queue<int> q;
  q.push(s);
  int f=0;
  while(!q.empty())
  {
    int u = q.front();
    q.pop();
    if(u==t) break;
    for(auto v: e[u])
    {
      if(!vis[v])
      {
		q.push(v);
		vis[v]=1;
        dist[v]=dist[u]+1;
      }
    }

  }
  if(dist[t]!=0 && dist[t]%3==0) {
  cout<<dist[t]/3;
  }
  else cout<<-1;
  return 0;
}
```