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)

#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
typedef long double ld;
const int inf=1e9+7;
const ll longinf=1LL<<60;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define F first
#define S second

const int mx=100010;
const ll mod=1e9+7;

typedef vector<vector<pair<ll,int>>> wgraph;
typedef pair<ll,int> pli;

vector<ll> bfs(wgraph v,int x){
  priority_queue<pli,vector<pli>,greater<pli>> q;
  vector<ll> dist(v.size(),longinf);
  dist[x]=0;
  q.push({0LL,x});

  while(q.size()){
    pli p=q.top(); q.pop();
    ll d=p.F, now=p.S;
    if(dist[now]<d){ continue; }
    for(auto to:v[now]){
      if(dist[to.S]>d+to.F){
        dist[to.S] = d+to.F;
        q.push({dist[to.S], to.S});
      }
    }
  }
  return dist;
}

int main(){
  int n,m;
  cin >> n >> m;
  wgraph v(mx*3);
  rep(i,m){
    int x,y,z=1;
    cin >> x >> y; x--; y--;
    v[x].emplace_back(z,y+mx);
    v[x+mx].emplace_back(z,y+mx+mx);
    v[x+mx+mx].emplace_back(z,y);
  }
  int s,t;
  cin >> s >> t; s--; t--;
  vector<ll> dist=bfs(v,s);
  ll ans = dist[t];
  if(ans==longinf){
    ans = -1;
  } else {
    ans/=3;
  }
  cout << ans << endl;
  return 0;
}