#include <bits/stdc++.h>

using namespace std;

int main() {
  int n,m; cin >> n >> m;
  vector<vector<int> > graph(n);
  int u,v,s,t;
  for(int i=0;i<m;++i){
    cin >> u >> v;
    graph[u-1].push_back(v-1);
  }
  cin >> s >> t;
  s--,t--;
  
  vector<vector<long long> > dist(n,vector<long long>(3,-1));
  dist[s][0]=0;
  queue<pair<int,int> > q;
  q.push(make_pair(s,0));
  while(!q.empty()){
    pair<int,int> now=q.front(); q.pop();
    int pto=(now.second+1)%3;
    for(auto to : graph[now.first]){
      if(dist[to][pto]!=-1) continue;
      dist[to][pto]=dist[now.first][now.second]+1;
      q.push(make_pair(to,pto));
    }
  }
  if(dist[t][0]==-1) cout << -1 << endl;
  else cout << dist[t][0]/3 << endl; 
  
  return 0;
}

### 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>

using namespace std;

int main() {
  int n,m; cin >> n >> m;
  vector<vector<int> > graph(n);
  int u,v,s,t;
  for(int i=0;i<m;++i){
    cin >> u >> v;
    graph[u-1].push_back(v-1);
  }
  cin >> s >> t;
  s--,t--;
  
  vector<vector<long long> > dist(n,vector<long long>(3,-1));
  dist[s][0]=0;
  queue<pair<int,int> > q;
  q.push(make_pair(s,0));
  while(!q.empty()){
    pair<int,int> now=q.front(); q.pop();
    int pto=(now.second+1)%3;
    for(auto to : graph[now.first]){
      if(dist[to][pto]!=-1) continue;
      dist[to][pto]=dist[now.first][now.second]+1;
      q.push(make_pair(to,pto));
    }
  }
  if(dist[t][0]==-1) cout << -1 << endl;
  else cout << dist[t][0]/3 << endl; 
  
  return 0;
}

```