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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
using ll = long long;
const int N = 1e5;

int n, m, s, t;

vector<int> conn[N];
int dist[N], ans = -1;
set<int> loc[N];

int main() {
	cin.sync_with_stdio(0); cin.tie(0);
	cin >> n >> m;
	for (int i = 0; i < m; ++i) {
		int a, b;
		cin >> a >> b;
		--a, --b;
		conn[b].push_back(a);
	}
	cin >> s >> t;
	--s, --t;
	queue<int> q;
	q.push(t);
	
	
	while (!q.empty()) {
		int cur = q.front();
		q.pop();
		
		queue<pair<int, int>> tq;
		tq.push({cur, 0});
		while (!tq.empty()) {
			pair<int, int> v = tq.front();
			tq.pop();
			if (v.second == 3) {
				if (!dist[v.first] && v.first != t) {
					dist[v.first] = dist[cur] + 1;
					q.push(v.first);
					if (v.first == s) {
						tq = {};
						q = {};
						break;
					}
				}
			} else {
				for (int u : conn[v.first]) {
					if (loc[u].insert(v.second + 1).second) tq.push({u, v.second + 1});
				}
			}
		}
	}
	
	dist[s] ? cout << dist[s] : cout << -1;
}