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
#include <bits/stdc++.h>
using namespace std;
using Graph = vector<vector<int>>;

int main () {
  
  // 入力
  
  int N, M;
  cin >> N >> M;
  
  Graph G(N);
  
  for (int i = 0; i < M; i++) {
    int a, b;
    cin >> a >> b;
    G[a - 1].push_back(b - 1);
  }
  
  int S, T;
  cin >> S >> T;
  S--, T--;
  
  // BFS
  
  vector<vector<int>> step(3, vector<int>(N, -3));
  queue<pair<int, int>> que;
  
  step[0][S] = 0;
  que.push({0, S});
  
  while (!que.empty()) {
    
    pair<int, int> p = que.front();
    que.pop();
    int mod = p.first;
    int x = p.second;
    
    for (int nx : G[x]) {
      if (step[(mod + 1) % 3][nx] == -3) {
        step[(mod + 1) % 3][nx] = step[mod][x] + 1;
        que.push({(mod + 1) % 3, nx});
      }
    }
    
  }
  
  // 出力
  
  cout << step[0][T] / 3 << endl;
  
}