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: Wrong Answer)

#include <bits/stdc++.h>
#include <unordered_set>
using namespace std;
#define ll long long
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define repeat(i,s,n) for(int (i)=s; (i)<(n); (i)++)
#define revrep(i,n) for(int (i)=(n)-1;i>=0; i--)

template<typename W>
// example const ll inf=1e15;
void dijkstra(int s, map<int,map<int,W>>& g, vector<W>& dist, const W inf) {
  vector<bool> visited(dist.size());
  fill(dist.begin(),dist.end(),inf);
  using P = pair<W,int>;
  priority_queue<P,vector<P>, greater<P>> pq;
  dist[s]=0;
  //visited[s]=true;
  pq.push(make_pair(0,s));
  while(!pq.empty()) {
    int v;
    W d;
    tie(d,v)=pq.top();
    pq.pop();
    if(visited[v]) continue;
    visited[v]=true;
    for(auto& uw : g[v]) {
      int u;
      W w;
      tie(u,w)=uw;
      if(visited[u]) continue;
      if(d+w<dist[u]) {
        dist[u]=d+w;
        pq.push(make_pair(dist[u],u));
      }
    }
  }
}


using G = map<int,map<int,int>>;
G make_kg(int n, vector<vector<int>> g) {
  G kg;
  repeat(v,1,n+1) {
    // move three times from s
    for(auto& u : g[v]) {
      kg[v][u+n]=1;
      kg[v+n][u+2*n]=1;
      kg[v+2*n][u]=1;
    }
  }

  return kg;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<setprecision(std::numeric_limits<float>::max_digits10);
  cout<<fixed;
  int n,m;
  cin>>n>>m;
  vector<vector<int>> g(n+1);
  rep(i,m) {
    int u,v;
    cin>>u>>v;
    g[u].push_back(v);
  }
  int s,t;
  cin>>s>>t;
  G kg = make_kg(n,g);
  vector<int> dist(3*n+1);
  const int inf=1e8;
  dijkstra<int>(s,kg,dist,inf);
  if(dist[t]==inf) {
    cout << -1 << endl;
  } else {
    cout << dist[t] << endl;
  }
  return 0;
}