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 <bits/stdc++.h>

template <typename InputIterator>
typename InputIterator::value_type summation(InputIterator first,
                                             InputIterator last) {
  return std::accumulate(first, last, typename InputIterator::value_type());
}
template <typename T>
std::istream &operator>>(std::istream &stream, std::vector<T> &v);
template <typename T1, typename T2>
std::istream &operator>>(std::istream &stream, std::pair<T1, T2> &p);
template <typename T>
std::istream &operator>>(std::istream &stream, std::vector<T> &v) {
  for (auto &i : v) {
    stream >> i;
  }
  return stream;
}
template <typename T1, typename T2>
std::istream &operator>>(std::istream &stream, std::pair<T1, T2> &p) {
  stream >> p.first >> p.second;
  return stream;
}

int main() {
  int64_t n, m;
  std::cin >> n >> m;
  std::vector<std::unordered_set<int64_t>> no(n);

  for (int64_t i = 0; i < m; i++) {
    int64_t u, v;
    std::cin >> u >> v;
    u--;
    v--;
    no[u].insert(v);
  }

  int64_t s, t;
  std::cin >> s >> t;
  s--;
  t--;

  std::vector<std::unordered_set<int64_t>> no2(n);
  std::vector<bool> f_no2(n, false);

  const auto get2 = [&](auto i) {
    if (!f_no2[i]) {
      for (const auto j : no[i]) {
        for (const auto k : no[j]) {
          no2[i].insert(k);
        }
      }
      f_no2[i] = true;
    }
  };

  std::vector<std::unordered_set<int64_t>> no3(n);
  std::vector<bool> f_no3(n, false);

  const auto get3 = [&](auto i) {
    if (!f_no3[i]) {
      get2(i);
      for (const auto j : no2[i]) {
        for (const auto k : no[j]) {
          no3[i].insert(k);
        }
      }
      f_no3[i] = true;
    }
  };

  std::vector<bool> flag2(n, false);
  std::deque<int64_t> q2;
  flag2[s] = true;
  for (const auto i : no[s]) {
    if (i == t) {
      flag2[i] = true;
      q2.clear();
      break;
    }
    q2.push_back(i);
  }
  while (!q2.empty()) {
    const auto i = q2.front();

    if (flag2[i]) {
      q2.pop_front();
      continue;
    }
    flag2[i] = true;

    for (const auto j : no[i]) {
      if (j == t) {
        flag2[j] = true;
        q2.clear();
        break;
      }
      if (!flag2[j]) {
        q2.push_back(j);
      }
    }
  }

  if (!flag2[t]) {
    std::cout << -1 << std::endl;
    return 0;
  }

  std::vector<bool> flag(n, false);
  std::deque<std::pair<int64_t, int64_t>> q;

  flag[s] = true;
  get3(s);
  for (const auto i : no3[s]) {
    if (i == t) {
      std::cout << 1 << std::endl;
      return 0;
    }
    q.push_back(std::make_pair(i, 1));
  }

  while (!q.empty()) {
    const auto v = q.front();

    if (flag[v.first]) {
      q.pop_front();
      continue;
    }
    flag[v.first] = true;

    get3(v.first);
    for (const auto i : no3[v.first]) {
      if (i == t) {
        std::cout << v.second + 1 << std::endl;
        return 0;
      }
      if (!flag[i]) {
        q.push_back(std::make_pair(i, v.second + 1));
      }
    }
  }
  std::cout << -1 << std::endl;
  return 0;
}