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>
using namespace std;
using ll = long long;
const int MAX = 1e5+5;
const ll INF = 1e13+7;

ll N,M,S,T;
vector<vector<ll> > node, nnode;
map<tuple<ll,ll,int>, bool> ex;

void bfs(ll s, ll key, ll n) {
	if(ex.count(make_tuple(s,key,n))) return;
	ex[make_tuple(s,key,n)] = true;
	if(n == 3) {
		nnode[s].push_back(key);
		ex[make_tuple(s,key,n)] = true;
		return;
	}
	for(int to=0;to<node[key].size();to++) {
		if(ex.count(make_tuple(s, node[key][to], n+1))) {
			continue;
		}
		bfs(s, node[key][to], n+1);
	}
}

ll dijkstra(ll s, ll e){
  priority_queue<pair<ll,ll> > PQ;
  vector<bool> done(N,false);
  vector<ll> d(N, INF);

  d[s] = 0;
  PQ.push(make_pair(0,s));

  while(!PQ.empty()){
    pair<ll ,ll> f = PQ.top(); PQ.pop();
    ll u = f.second;
    done[u] = true;

    for(int j=0;j<nnode[u].size();j++){
      int v = nnode[u][j];
      if(done[v] == true) continue;
      if(d[v] > d[u] + 1){
        d[v] = d[u] + 1;
        PQ.push(make_pair(d[v] * (-1), v));
      }
    }
  }
  return (d[e] == INF) ? -1 : d[e];
}

signed
main(){
	cin >> N >> M;
	node = vector<vector<ll> >(N+4);
	nnode = vector<vector<ll> >(N+4);
	for(int i=0;i<M;i++) {
		ll u,v;
		cin >> u >> v;
		u--, v--;
		node[u].push_back(v);
	}
	cin >> S >> T;
	S--, T--;

	for(int i=0;i<N;i++) {
		bfs(i,i,0);
	}
	cout << dijkstra(S,T) << endl;
}