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: Accepted)

#pragma GCC optimize ("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <bits/stdc++.h>
using namespace std;
using Int = long long;
typedef pair<int,int> P; typedef pair<Int,Int> Pl;
const int mod = 1e9+7;
const Int Inf = 1e17;
#define END {cout<<ans<<'\n'; return 0;}
#define ALL(v) (v).begin(),(v).end()
#define fi first
#define se second
#define mp make_pair
#define Pr(t) priority_queue<t>
#define gPr(t) priority_queue<t,vector<t>,greater<t>>
#define V(t) vector<t>
#define rep(i,n) for(int(i)=0; i<n; i++)
#define rer(i,l,r) for(int i=l; i<=r; i++)
#define gnr(i,l,r) for(int i=r; i>=l; i--)
#define eb emplace_back
#define pri1(a) cout<<(a)<<'\n'
#define pri2(a,n) {rep(pri2i,n-1)cout<<a[pri2i]<<' '; if(n>1)cout<<a[n-1]<<'\n';}
#define prip(p) cout<<p.fi<<' '<<p.se<<'\n'
template<class T> inline bool cmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool cmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template< class T >
struct edge {
  int fr, to;
  T w;

  edge(int to, T w) : fr(-1), to(to), w(w) {}

  edge(int fr, int to, T w) : fr(fr), to(to), w(w) {}

  edge &operator=(const int &x) {
    to = x;
    return *this;
  }

  operator int() const { return to; }
};

template< typename T >
using Edges = vector< edge< T > >;
template< typename T >
using WeightedGraph = vector< Edges< T > >;
using UnWeightedGraph = vector< vector< int > >;
template< typename T >
using Matrix = vector< vector< T > >;

template< class T >
vector< T > dijkstra(WeightedGraph< T > &g, int s) {
  const auto INF = numeric_limits< T >::max();
  vector< T > dist(g.size(), INF);

  using Pi = pair< T, int >;
  priority_queue< Pi, vector< Pi >, greater< Pi > > que;
  dist[s] = 0;
  que.emplace(dist[s], s);
  while(!que.empty()) {
    T cost;
    int idx;
    tie(cost, idx) = que.top();
    que.pop();
    if(dist[idx] < cost) continue;
    for(auto &e : g[idx]) {
      auto next_cost = cost + e.w;
      if(dist[e.to] <= next_cost) continue;
      dist[e.to] = next_cost;
      que.emplace(dist[e.to], e.to);
    }
  }
  return dist;
}

int n,m,_,x,y,q;
string s,sb;
bool ok;
char cf,cb;

int main(){
  cin.tie(nullptr); ios::sync_with_stdio(0);

  cin>>n>>m; WeightedGraph<int>g(n*3);
  rep(i,m){
    cin>>x>>y; x--; y--;
    g[x].eb(y+n,1);
    g[x+n].eb(y+n*2,1);
    g[x+n*2].eb(y,1);
  }
  cin>>x>>y; x--,y--;
  auto ans=dijkstra(g,x);
  pri1(ans[y]>1e9?-1:ans[y]/3);
}