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)

// OURL: https://atcoder.jp/contests/abc132/tasks/abc132_e
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
#define REP(i,n) for(int i=0,i##_len=(n);i<i##_len;++i)
#define RREP(i,a,n) for(int i=(a),i##_len=(n);i<i##_len;++i)
#define MAT1D(V,T,a) vector<T> V (a)
#define MAT2D(V,T,a,b) vector<vector<T>> V (a, vector<T>(b))
#define MAT3D(V,T,a,b,c) vector<vector<vector<T>>> V (a, vector<vector<T>>(b, vector<T>(c)))

// http://beet-aizu.hatenablog.com/entry/2018/04/08/145516
template<typename T>
vector<T> make_v(size_t a){return vector<T>(a);}
template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){ return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...)); }

template<typename T,typename V>
typename enable_if<is_class<T>::value==0>::type
fill_v(T &t,const V &v){t=v;}
template<typename T,typename V>
typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t,const V &v){ for(auto &e:t) fill_v(e,v); }

template <typename A>
string to_string(A v) { bool first = true; string res = "{"; for (const auto &x : v) { if (!first) { res += ", "; } first = false; res += to_string(x); } res += "}"; return res; }

string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string) s); }
string to_string(bool b) { return (b ? "true" : "false"); }
string to_string(vector<bool>::reference b) { return to_string(bool(b)); }
template <typename A, typename B>
string to_string(pair<A, B> p) { return "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; }
template <typename A>
string to_string(vector<vector<A>> v) { string res; for(const auto &x : v) {res+="\n ";res += to_string(x);} return res;}

void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) { cerr << " " << to_string(H); debug_out(T...); }

#ifdef LOCAL
#define debug(...) cerr << "[L" << to_string(__LINE__) << ":" <<#__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

inline int toidx(int v) {
  return (v-1)*3;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N,M;
  cin >> N >> M;

  auto edge = make_v<set<int>>(N*3);
  REP(i,M) {
    int u1, u2;
    cin >> u1 >> u2;
    u1 = toidx(u1);
    u2 = toidx(u2);
    edge[u1].insert(u2+1);
    edge[u1+1].insert(u2+2);
    edge[u1+2].insert(u2);
  }
  int S,T;
  cin >> S >> T;
  S = toidx(S);
  T = toidx(T);
  debug(S,T);
  auto min_cnt = make_v<int>(N*3);
  fill_v(min_cnt, INT_MAX);
  set<int> vert;
  auto done = make_v<bool>(N*3);
  // set<int> done;
  min_cnt[S] = 0;
  bool fin = false;
  // vert.insert(S);
  deque<int> que;
  que.push_back(S);
  while (que.size() > 0) {
    int v = que.front();que.pop_front();
    // set<int> tmp;
    // for (auto& v: vert) {
    //   // if (done[v]) continue;

      for (auto &vv : edge[v]) {
        // tmp.insert(vv);
        min_cnt[vv] = min(min_cnt[vv], min_cnt[v]+1);
        if (vv == T) {
          fin=true; break;
        }
        if (!done[vv]) que.push_back(vv);
      }
      done[v] = true;
      if (fin) break;
  // }
    // vert = move(tmp);
    // for (auto &vv: tmp) {
    //   vert.insert(vv);
    // }
    // if (fin) break;
  }
  // debug(done,vert);
  debug(min_cnt);
  if (fin) {
    cout << min_cnt[T] / 3 << endl;
  } else {
    cout << -1 << endl;
  }
  return 0;
}