#include<bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int inf = 1001001001;
int main(){
  int N,M,S,T,i,u,v;scanf("%d%d",&N,&M);
  vector<int> G[N];
  for(i=0;i<M;i++){
    scanf("%d%d",&u,&v);u--;v--;
    G[u].push_back(v);
  }
  scanf("%d%d",&S,&T);S--;T--;
  int D[N][3];
  for(i=0;i<N;i++) fill(D[i],D[i]+3,inf);
  D[S][0] = 0;
  queue<pair<int,int>> q;q.push(make_pair(S,0));
  int fn,fd;
  while(!q.empty()){
    fn = q.front().first;fd = q.front().second;q.pop();
    for(i=0;i<G[fn].size();i++){
      if(D[G[fn][i]][(fd+1)%3] == inf){
        D[G[fn][i]][(fd+1)%3] = D[fn][fd]+1;
        q.push(make_pair(G[fn][i],(fd+1)%3));
      }
    }
  }
  if(D[T][0] == inf) printf("-1\n");
  else printf("%d\n",D[T][0]/3);
}

### 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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int inf = 1001001001;
int main(){
  int N,M,S,T,i,u,v;scanf("%d%d",&N,&M);
  vector<int> G[N];
  for(i=0;i<M;i++){
    scanf("%d%d",&u,&v);u--;v--;
    G[u].push_back(v);
  }
  scanf("%d%d",&S,&T);S--;T--;
  int D[N][3];
  for(i=0;i<N;i++) fill(D[i],D[i]+3,inf);
  D[S][0] = 0;
  queue<pair<int,int>> q;q.push(make_pair(S,0));
  int fn,fd;
  while(!q.empty()){
    fn = q.front().first;fd = q.front().second;q.pop();
    for(i=0;i<G[fn].size();i++){
      if(D[G[fn][i]][(fd+1)%3] == inf){
        D[G[fn][i]][(fd+1)%3] = D[fn][fd]+1;
        q.push(make_pair(G[fn][i],(fd+1)%3));
      }
    }
  }
  if(D[T][0] == inf) printf("-1\n");
  else printf("%d\n",D[T][0]/3);
}

```