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)

use io::*;
use std::collections::BinaryHeap;

fn main() {
	let con = read_string();
	let mut sc = Scanner::new(&con);

	let n: usize = sc.next();
	let m: usize = sc.next();

	let inf: i32 = std::i32::MAX / 2;

	let mut g: Vec<Vec<usize>> = vec![Vec::new(); n];

	for _ in 0..m {
		let mut a: usize = sc.next();
		let mut b: usize = sc.next();
		a -= 1;
		b -= 1;

		g[a].push(b);
	}

	let mut s: usize = sc.next();
	let mut t: usize = sc.next();
	s -= 1;
	t -= 1;

	let mut dist: Vec<Vec<i32>> = vec![vec![inf; n]; 3];

	let mut pq = BinaryHeap::new();

	dist[0][s] = 0;
	pq.push((-dist[0][s], (0, s)));

	while let Some((d, (i, v))) = pq.pop() {
		let d = -d;

		if dist[i][v] < d {
			continue;
		}

		for u in &g[v] {
			let j: usize = (i + 1) % 3;
			let u: usize = *u;

			if dist[i][v] + 1 < dist[j][u] {
				dist[j][u] = dist[i][v] + 1;
				pq.push((-dist[j][u], (j, u)));
			}
		}
	}

	if dist[0][t] == inf {
		println!("-1");
		std::process::exit(0);
	}

	println!("{}", dist[0][t] / 3);
}

pub mod io {
	use std;
	use std::str::FromStr;

	pub struct Scanner<'a> {
		iter: std::str::SplitWhitespace<'a>,
	}

	impl<'a> Scanner<'a> {
		pub fn new(s: &'a str) -> Scanner<'a> {
			Scanner {
				iter: s.split_whitespace(),
			}
		}

		pub fn next<T: FromStr>(&mut self) -> T {
			let s = self.iter.next().unwrap();
			if let Ok(v) = s.parse::<T>() {
				v
			} else {
				panic!("Parse error")
			}
		}

		pub fn next_vec_len<T: FromStr>(&mut self) -> Vec<T> {
			let n: usize = self.next();
			self.next_vec(n)
		}

		pub fn next_vec<T: FromStr>(&mut self, n: usize) -> Vec<T> {
			(0..n).map(|_| self.next()).collect()
		}
	}

	pub fn read_string() -> String {
		use std::io::Read;

		let mut s = String::new();
		std::io::stdin().read_to_string(&mut s).unwrap();
		s
	}

	pub fn read_line() -> String {
		let mut s = String::new();
		std::io::stdin().read_line(&mut s).unwrap();
		s.trim_right().to_owned()
	}
}