#include <vector>
#include <queue>
#include <iostream>
#include <queue>
#include <utility>
#include <tuple>
using namespace std;
constexpr int INF = 1e9;

int main(){
  int N, M;
  cin >> N >> M;
  vector<vector<int>> G(N);
  for(int i = 0; i < M; ++i){
    int u, v;
    cin >> u >> v;
    --u,--v;
    G[u].push_back(v);
  }
  int s, t;
  cin >> s >> t;
  --s,--t;
  vector<vector<int>> D(N,vector<int>(3,INF));
  D[s][0] = 0;
  queue<pair<int,int>> Q;
  Q.emplace(s,0);
  while(Q.size()){
    int v, d;
    tie(v,d) = Q.front();
    Q.pop();
    int c = D[v][d] + (d == 0);
    int d_ = (d+1)%3;
    for(auto v_ : G[v]){
      if(D[v_][d_] <= c) continue;
      D[v_][d_] = c;
      Q.emplace(v_,d_);
    }
  }
  if(D[t][0] >= INF){
    cout << -1 << endl;
    return 0;
  }
  cout << D[t][0] << endl;
}

### Prompt

Generate a Cpp solution 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 <vector>
#include <queue>
#include <iostream>
#include <queue>
#include <utility>
#include <tuple>
using namespace std;
constexpr int INF = 1e9;

int main(){
  int N, M;
  cin >> N >> M;
  vector<vector<int>> G(N);
  for(int i = 0; i < M; ++i){
    int u, v;
    cin >> u >> v;
    --u,--v;
    G[u].push_back(v);
  }
  int s, t;
  cin >> s >> t;
  --s,--t;
  vector<vector<int>> D(N,vector<int>(3,INF));
  D[s][0] = 0;
  queue<pair<int,int>> Q;
  Q.emplace(s,0);
  while(Q.size()){
    int v, d;
    tie(v,d) = Q.front();
    Q.pop();
    int c = D[v][d] + (d == 0);
    int d_ = (d+1)%3;
    for(auto v_ : G[v]){
      if(D[v_][d_] <= c) continue;
      D[v_][d_] = c;
      Q.emplace(v_,d_);
    }
  }
  if(D[t][0] >= INF){
    cout << -1 << endl;
    return 0;
  }
  cout << D[t][0] << endl;
}

```