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() {
  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.at(u).push_back(v);
  }
  int S, T; cin >> S >> T;
  S--; T--;
  const int INF = 1000000007;
  vector<vector<int>> D(N, vector<int>(3, INF));
  D.at(S).at(0) = 0;
  queue<pair<int, int>> q;
  q.push(make_pair(S, 0));
  while (!q.empty()) {
    int cur = q.front().first;
    int d = q.front().second;
    q.pop();
    int nd = (d + 1) % 3;
    for (int nx : G.at(cur)) {
      if (D.at(nx).at(nd) != INF) continue;
      D.at(nx).at(nd) = D.at(cur).at(d) + 1;
      q.push(make_pair(nx, nd));
    }
  }
  if (D.at(T).at(0) != INF) cout << D.at(T).at(0) / 3 << endl;
  else cout << -1 << endl;
}