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 std::io::{self, BufRead};
use std::mem::swap;
use std::collections::BTreeSet;

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let xs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
  let (n, m) = (xs[0], xs[1]);
  let mut es: Vec<Vec<usize>> = vec![vec![]; n];
  for _ in 0..m {
    let xs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
    let (u, v) = (xs[0] - 1, xs[1] - 1);
    es[u].push(v);
  }
  let xs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
  let (s, t) = (xs[0] - 1, xs[1] - 1);
  let mut dss: Vec<Vec<usize>> = vec![vec![usize::max_value(); 3]; n];
  dss[s][0] = 0;
  let mut nexts: BTreeSet<usize> = BTreeSet::new();
  nexts.insert(s);
  let mut step: usize = 1;
  while !nexts.is_empty() {
    let mut temp: BTreeSet<usize> =BTreeSet::new();
    let round = (step - 1) / 3 + 1;
    let phase = step % 3;
    for &u in &nexts {
      for &v in &es[u] {
        if round < dss[v][phase] {
          dss[v][phase] = round;
          temp.insert(v);
        }
      }
    }
    swap(&mut nexts, &mut temp);
    step += 1;
  }
  println!("{}", if dss[t][0] < usize::max_value() { dss[t][0] as i32 } else { -1 });
}