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 Rust (1.15.1) (judge verdict: Accepted)

fn main() {
	use std::io::Read;
	let mut s = String::new();
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let m: usize = s.next().unwrap().parse().unwrap();
	let mut edges = vec![vec![]; n * 3];
	for _ in 0..m {
		let u = (s.next().unwrap().parse::<usize>().unwrap() - 1) * 3;
		let v = (s.next().unwrap().parse::<usize>().unwrap() - 1) * 3;
		edges[u].push(v + 1);
		edges[u + 1].push(v + 2);
		edges[u + 2].push(v);
	}
	let src = (s.next().unwrap().parse::<usize>().unwrap() - 1) * 3;
	let dst = (s.next().unwrap().parse::<usize>().unwrap() - 1) * 3;
	let mut dist = vec![std::i32::MAX; n * 3];
	dist[src] = 0;
	let mut q = std::collections::BinaryHeap::new();
	q.push((0, src));
	while let Some((cost, v)) = q.pop() {
		let cost = -cost;
		if cost > dist[v] {
			continue;
		}
		for &e in edges[v].iter() {
			let next_cost = cost + 1;
			if next_cost < dist[e] {
				dist[e] = next_cost;
				q.push((-next_cost, e));
			}
		}
	}
	if dist[dst] == std::i32::MAX {
		println!("-1");
	} else {
		println!("{}", dist[dst] / 3);
	}
}