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 lli = int_fast64_t;
using ulli = uint_fast64_t;
#define debug(x) cerr << #x << " : " << x << endl
#define debugv(x) cerr << #x << " : {";for(auto value_of_x:x){ cerr << value_of_x << ", ";}cerr << endl
#define debugst(x) auto xt = x;cerr << #x << " : {";for(int i=0;i<x.size();i++){ cerr << xt.top() << ", ";xt.pop();}cerr << endl
#define debugvp(x) cerr << #x << " : {";for(auto value_of_x:x){ cerr << value_of_x.first << " : " << value_of_x.second << ", ";}cerr << endl

const lli MOD = 1000000007;
const lli LIMIT = 1000000000000000000 * 5;
const int tenfive = 100000;

using graph = vector<vector<int>>;
using st = pair<int, int>;
int main() {
	int N, M;
	cin >> N >> M;
	graph G(N);
	for (int i = 0; i < M; i++) {
		int x, y;
		cin >> x >> y;
		G[x-1].push_back(y-1);
	}
	int S, T;
	cin >> S >> T;
	S--; T--;
	vector<bool> visited(N, false);
	queue<st> q;
	q.push({ S,0 });
	while (!q.empty()) {
		st now = q.front(); q.pop();
		if (now.first == T) {
			cout << now.second << endl;
			return 0;
		}
		now.second++;
		if (visited[now.first])continue;
		visited[now.first] = true;
		set<int> start = { now.first };
		for (int i = 0; i < 3; i++) {
			set<int> next;
			for (int j:start) {
				for (int k:G[j]) {
					next.insert(k);
				}
			}
			start = move(next);
		}
		for (int j : start) {
			q.push({ j,now.second});
		}
	}
	cout << -1 << endl;
	return 0;
}