Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

4 4
1 2
2 3
3 4
4 1
1 3

Sample Output 1

2

Ken can reach Vertex 3 from Vertex 1 in two ken-ken-pa, as follows: 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 in the first ken-ken-pa, then 4 \rightarrow 1 \rightarrow 2 \rightarrow 3 in the second ken-ken-pa. This is the minimum number of ken-ken-pa needed.

Sample Input 2

3 3
1 2
2 3
3 1
1 2

Sample Output 2

-1

Any number of ken-ken-pa will bring Ken back to Vertex 1, so he cannot reach Vertex 2, though he can pass through it in the middle of a ken-ken-pa.

Sample Input 3

2 0
1 2

Sample Output 3

-1

Vertex S and Vertex T may be disconnected.

Sample Input 4

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6

Sample Output 4

2

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <unordered_set>
#include <queue>

using namespace std;

int bfs(vector<unordered_set<int> >& G, int n, int s, int t) {
  if (s == t) {
    return 0;
  }
  vector<bool> vis(n);
  queue<pair<int, int> > q;
  q.emplace(s, 0);
  while (!q.empty()) {
    pair<int, int> p = q.front();
    int i = p.first;
    int a = p.second;
    q.pop();
    for (int j : G[i]) {
      if (!vis[j]) {
        if (j == t) {
          return a + 1;
        }
        vis[j] = true;
        q.emplace(j, a + 1);
      }
    }
  }
  return -1;
}

int main() {
  int n, m;
  cin >> n >> m;
  vector<unordered_set<int> > G1(n);
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    G1[u].insert(v);
  }
  int s, t;
  cin >> s >> t;
  s--;
  t--;
  vector<unordered_set<int> > G2(n);
  for (int i = 0; i < n; i++) {
    for (int j : G1[i]) {
      for (int k : G1[j]) {
        G2[i].insert(k);
      }
    }
  }
  vector<unordered_set<int> > G3(n);
  for (int i = 0; i < n; i++) {
    for (int j : G1[i]) {
      for (int k : G2[j]) {
        G3[i].insert(k);
      }
    }
  }
  int ans = bfs(G3, n, s, t);
  cout << ans << endl;
  return 0;
}