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: Wrong Answer)

#ifdef __GNUC__
#pragma GCC diagnostic ignored "-Wunused-result"
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif

#include <algorithm>
#include <cstring>
#include <iostream>
#include <sstream>
#include <queue>
#include <vector>

using namespace std;

static const int INF = 1 << 30;

void dfs(int n, int d, vector<vector<int>> &e, vector<vector<int>>& vis) {
	int& r = vis[n][d % 3];
	if (r > d) {
		r = d;
		for (int next : e[n]) {
			dfs(next, d + 1, e, vis);
		}
	}
}

void solve(int N, int M, std::vector<int>& u, std::vector<int>& v, int S, int T) {
	vector<vector<int>> vis(N, vector<int>(3, INF));
	vector<vector<int>> e(N);
	for (int i = 0; i < M; i++) {
		e[u[i]].emplace_back(v[i]);
	}
	dfs(S, 0, e, vis);
	int ans = vis[T][2];
	cout << (ans >= INF ? -1 : ans) << endl;
}

int main() {
	int N, M, S, T;
	std::cin >> N >> M;
	std::vector<int> u(M), v(M);
	for (int i = 0; i < M; i++) {
		std::cin >> u[i] >> v[i];
		u[i]--, v[i]--;
	}
	std::cin >> S >> T;
	--S, --T;
	solve(N, M, u, v, S, T);
	return 0;
}