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;

int main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  int n, m, x, y, s, t;
  vector <int> a[100100];
  cin >> n >> m;
  while (m--)
  {
    cin >> x >> y;
    a[x].push_back(y);
  }

  cin >> s >> t;
  queue < pair<int, int> > q;
  q.push({s, 0});
  int dist[100100][3];
  memset(dist, -1, sizeof dist);
  dist[s][0] = 0;

  while (!q.empty())
  {
    auto u = q.front();
    q.pop();
    int x = u.first, z = u.second;
    for (int y : a[x])
      if (dist[y][(z + 1) % 3] < 0)
      {
        dist[y][(z + 1) % 3] = dist[x][z] + 1;
        q.push({y, (z + 1) % 3});
      }
  }

  cout << (dist[t][0] >= 0 ? dist[t][0] / 3 : -1) << endl;
}
