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::collections::HashMap;
use std::collections::VecDeque;
use std::io::{self, Read};
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = io::stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char) 
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}
type Point = (usize, usize);
type Graph = HashMap<Point, Vec<Point>>;
type Todo = VecDeque<Point>;

fn dijkstra(g: &Graph, v0: Point, v1: Point) -> i64 {
    let mut pot: HashMap<Point, i64> = HashMap::new();
    pot.insert(v0, 0);
    let mut todo: Todo = VecDeque::new();
    todo.push_back(v0);
    while !todo.is_empty() {
        let cur = todo.front().unwrap().clone();
        todo.pop_front();
        // eprintln!("cur:{},{}, pot:{}", cur.0, cur.1, *pot.get(&cur).unwrap());
        match g.get(&cur) {
            Some(ws) =>
                for w in ws {
                    if !pot.contains_key(w) {
                        let dist = pot.get(&cur).unwrap() + 1;
                        pot.insert(*w, dist);
                        todo.push_back(*w);
                        if *w == v1 {
                            return dist / 3;
                        }
                    }
                },
            None => (),
        }
    }
    -1
}

fn main() {
    let _: usize = read();
    let m: usize = read();
    let mut g: Graph = HashMap::new();
    for _ in 0..m {
        let p: usize = read::<usize>() - 1;
        let q: usize = read::<usize>() - 1;
        for i in 0..3 {
            g.entry((p, i)).or_insert(Vec::new()).push((q, (i + 1) % 3));
        }
    }
    let s: usize = read::<usize>() - 1;
    let t: usize = read::<usize>() - 1;
    let ans = dijkstra(&g, (s, 0), (t, 0));
    println!("{}", ans);
}