#include <bits/stdc++.h>

using ll = long long;
using namespace std;
const int INFint = 1e9+1;
const ll INFll = (ll)1e18+1;
ll MOD=1e9+7;

int main(){
  int N,M;
  vector<int> G[100000];
  int s,t;
  vector<vector<ll>> dp(100000,vector<ll>(3,INFll));
  cin>>N>>M;
  for(int i(0);i<M;i++){
    int u,v;
    cin>>u>>v;
    u--;v--;
    G[u].push_back(v);
  }
  cin>>s>>t;
  s--;t--;
  dp[s][0] = 0;
  queue<pair<int,int>> q;
  q.push({s,0});
  while(!q.empty()){
    pair<int,int> now = q.front();q.pop();
    for(auto next:G[now.first]){
      if(dp[next][(now.second+1)%3] == INFll){
        dp[next][(now.second+1)%3] = now.second + 1;
        q.push({next,now.second+1});
      }
    }
  }
  if(dp[t][0] != INFll){
    cout << dp[t][0]/3 << endl;
  }else{
    cout << -1 << endl;
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 ll = long long;
using namespace std;
const int INFint = 1e9+1;
const ll INFll = (ll)1e18+1;
ll MOD=1e9+7;

int main(){
  int N,M;
  vector<int> G[100000];
  int s,t;
  vector<vector<ll>> dp(100000,vector<ll>(3,INFll));
  cin>>N>>M;
  for(int i(0);i<M;i++){
    int u,v;
    cin>>u>>v;
    u--;v--;
    G[u].push_back(v);
  }
  cin>>s>>t;
  s--;t--;
  dp[s][0] = 0;
  queue<pair<int,int>> q;
  q.push({s,0});
  while(!q.empty()){
    pair<int,int> now = q.front();q.pop();
    for(auto next:G[now.first]){
      if(dp[next][(now.second+1)%3] == INFll){
        dp[next][(now.second+1)%3] = now.second + 1;
        q.push({next,now.second+1});
      }
    }
  }
  if(dp[t][0] != INFll){
    cout << dp[t][0]/3 << endl;
  }else{
    cout << -1 << endl;
  }
  return 0;
}

```