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)

macro_rules ! input { ( source = $ s : expr , $ ( $ r : tt ) * ) => { let mut iter = $ s . split_whitespace ( ) ; input_inner ! { iter , $ ( $ r ) * } } ; ( $ ( $ r : tt ) * ) => { let s = { use std :: io :: Read ; let mut s = String :: new ( ) ; std :: io :: stdin ( ) . read_to_string ( & mut s ) . unwrap ( ) ; s } ; let mut iter = s . split_whitespace ( ) ; input_inner ! { iter , $ ( $ r ) * } } ; }
macro_rules ! input_inner { ( $ iter : expr ) => { } ; ( $ iter : expr , ) => { } ; ( $ iter : expr , mut $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let mut $ var = read_value ! ( $ iter , $ t ) ; input_inner ! { $ iter $ ( $ r ) * } } ; ( $ iter : expr , mut $ var : ident $ ( $ r : tt ) * ) => { let mut $ var = $ iter . next ( ) . unwrap ( ) . parse ::< usize > ( ) . unwrap ( ) ; input_inner ! { $ iter $ ( $ r ) * } } ; ( $ iter : expr , $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let $ var = read_value ! ( $ iter , $ t ) ; input_inner ! { $ iter $ ( $ r ) * } } ; ( $ iter : expr , $ var : ident $ ( $ r : tt ) * ) => { let $ var = $ iter . next ( ) . unwrap ( ) . parse ::< usize > ( ) . unwrap ( ) ; input_inner ! { $ iter $ ( $ r ) * } } ; }
macro_rules ! read_value { ( $ iter : expr , ( $ ( $ t : tt ) ,* ) ) => { ( $ ( read_value ! ( $ iter , $ t ) ) ,* ) } ; ( $ iter : expr , [ $ t : tt ; $ len : expr ] ) => { ( 0 ..$ len ) . map ( | _ | read_value ! ( $ iter , $ t ) ) . collect ::< Vec < _ >> ( ) } ; ( $ iter : expr , chars ) => { read_value ! ( $ iter , String ) . chars ( ) . collect ::< Vec < char >> ( ) } ; ( $ iter : expr , usize1 ) => { read_value ! ( $ iter , usize ) - 1 } ; ( $ iter : expr , $ t : ty ) => { $ iter . next ( ) . unwrap ( ) . parse ::<$ t > ( ) . unwrap ( ) } ; }
use std::collections::VecDeque;
#[derive(Debug)]
struct DirectedAcyclicGraph {
    graph: Vec<Vec<usize>>,
    cost: Vec<Option<usize>>,
}
impl DirectedAcyclicGraph {
    fn new(n: usize) -> DirectedAcyclicGraph {
        DirectedAcyclicGraph {
            graph: vec![vec![]; n],
            cost: vec![None; n],
        }
    }
    fn add_edge(&mut self, u: usize, v: usize) {
        self.graph[u].push(v)
    }
    fn bfs(&mut self, s: usize, t: usize) -> Option<usize> {
        self.cost = vec![None; self.graph.len()];
        let mut deq = VecDeque::new();
        self.cost[s] = Some(0);
        deq.push_back(s);
        while let Some(u) = deq.pop_front() {
            if u == t {
                return self.cost[u];
            }
            let d = self.cost[u].unwrap();
            for &v in &self.graph[u] {
                if self.cost[v].is_none() {
                    self.cost[v] = Some(d + 1);
                    deq.push_back(v);
                }
            }
        }
        None
    }
}
fn main() {
    input! { n, m, uv: [(usize1, usize1); m], s: usize1, t: usize1 };
    let mut dag = DirectedAcyclicGraph::new(3 * n);
    for &(u, v) in &uv {
        dag.add_edge(u, v + n);
        dag.add_edge(u + n, v + n * 2);
        dag.add_edge(u + n * 2, v);
    }
    println!("{}", dag.bfs(s, t).map(|x| x as isize / 3).unwrap_or(-1));
}