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)

// E - Hopscotch Addict
#include <bits/stdc++.h>

using namespace std;

void dfs(const vector<vector<int > > &g, int u, int v, int c, set<int> &vi)
{
	if (c == 0) {
		vi.insert(v);
		return;
	}
	for (int i = 0; i < g[v].size(); i++)
		dfs(g, u, g[v][i], c - 1, vi);
}

int main(int argc, char *argv[])
{
	ios::sync_with_stdio(false); cin.tie(0);
	int n, m, s, t;
	cin >> n >> m;
	vector<vector<int > > g(n), h(n);
	for (int i = 0; i < m; i++) {
		int u, v;
		cin >> u >> v;
		u--; v--;
		g[u].push_back(v);
	}
	cin >> s >> t;
	s--; t--;

	for (int i = 0; i < n; i++) {
		set<int> vi;
		dfs(g, i, i, 3, vi);
		for (set<int>::iterator it = vi.begin(); it != vi.end(); ++it)
			if (*it != i) h[i].push_back(*it);
	}

	vector<int> vi(n, -1);
	queue<pair<int, int> > q;
	for (q.push(make_pair(s, 0)); !q.empty(); q.pop()) {
		int u = q.front().first, d = q.front().second;
		vi[u] = d;
		for (int i = 0; i < h[u].size(); i++) {
			if (vi[h[u][i]] < 0) {
				vi[h[u][i]] = d + 1;
				q.push(make_pair(h[u][i], d + 1));
			}
		}
	}

	cout << vi[t] << endl;

	return 0;
}