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>
#define rep(i,n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;


int main()
{
  int n, m;
  cin >> n >> m;
  vector<vector<int>> g(n);
  rep(i, m)
  {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
  }
  int s, t;
  cin >> s >> t;
  s--;
  t--;
  vector<vector<bool>> chk(n, vector<bool>(3, false));
  queue<pair<int, int>> q;
  q.emplace(0, s);
  while (!q.empty())
  {
    auto p = q.front();
    q.pop();
    int count = p.first;
    int crt = p.second;
    if (chk[crt][count%3])
      continue;
    chk[crt][count%3] = true;
    if (count % 3 == 0 && crt == t)
    {
      cout << count / 3 << endl;
      return 0;
    }
    for (int i : g[crt])
      q.emplace(count + 1, i);
  }
  cout << -1 << endl;
  return 0;
}