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>
#include <assert.h>
using namespace std;
using i64 = long long;

int main()
{
  i64 n, m;
  cin >> n >> m;
  vector<i64> edge[n * 3];
  for (i64 i = 0; i < m; i++)
  {
    i64 u, v;
    cin >> u >> v;
    u--;
    v--;
    edge[u].push_back(n + v);
    edge[n + u].push_back(n * 2 + v);
    edge[n * 2 + u].push_back(v);
  }
  i64 s, t;
  cin >> s >> t;
  s--;
  t--;
  vector<i64> cost(n * 3, 1e18);
  queue<i64> que;
  cost[s] = 0;
  que.push(s);
  while (que.size())
  {
    i64 p = que.front();
    que.pop();
    for (i64 i : edge[p])
      if (cost[p] + 1 < cost[i])
      {
        cost[i] = cost[p] + 1;
        que.push(i);
      }
  }
  if (cost[t] == 1e18)
    cout << -1 << endl;
  else
    cout << cost[t] / 3 << endl;
  return 0;
}
