#include<bits/stdc++.h>
using namespace std;
int main(){
  cin.tie(0);
  ios::sync_with_stdio(0);
  int n,m;
  cin>>n>>m;
  vector<vector<int> > G(n);
  for(int i=0;i<m;i++){
    int x,y;
    cin>>x>>y;
    x--;y--;
    G[x].emplace_back(y);
  }
  int s,t;
  cin>>s>>t;
  s--;t--;
  vector<vector<int>> dp(3,vector<int>(n,-1));
  using P = pair<int, int>;
  queue<P> que;
  dp[0][s]=0;
  que.emplace(0,s);
  while(!que.empty()){
    int d,v;
    tie(d,v)=que.front();que.pop();//帅！
    int nd=(d+1)%3;
    for(int u:G[v]){
      if(~dp[nd][u]) continue;
      dp[nd][u]=dp[d][v]+1;
      que.emplace(nd,u);
    }
  }
  if(~dp[0][t]) dp[0][t]/=3;
  cout<<dp[0][t]<<endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int main(){
  cin.tie(0);
  ios::sync_with_stdio(0);
  int n,m;
  cin>>n>>m;
  vector<vector<int> > G(n);
  for(int i=0;i<m;i++){
    int x,y;
    cin>>x>>y;
    x--;y--;
    G[x].emplace_back(y);
  }
  int s,t;
  cin>>s>>t;
  s--;t--;
  vector<vector<int>> dp(3,vector<int>(n,-1));
  using P = pair<int, int>;
  queue<P> que;
  dp[0][s]=0;
  que.emplace(0,s);
  while(!que.empty()){
    int d,v;
    tie(d,v)=que.front();que.pop();//帅！
    int nd=(d+1)%3;
    for(int u:G[v]){
      if(~dp[nd][u]) continue;
      dp[nd][u]=dp[d][v]+1;
      que.emplace(nd,u);
    }
  }
  if(~dp[0][t]) dp[0][t]/=3;
  cout<<dp[0][t]<<endl;
  return 0;
}

```