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::VecDeque;

use std::io::*;
use std::str::FromStr;

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let stdin = 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 Weight = i32;
#[allow(dead_code)]
#[derive(Clone, Debug, Eq, PartialEq)]
struct Edge {
    index: usize,
    src: usize,
    dest: usize,
    weight: Weight,
}
#[allow(dead_code)]
impl Edge {
    fn new(index: usize, src: usize, dest: usize, weight: Weight) -> Edge {
        Edge {
            index: index,
            src: src,
            dest: dest,
            weight: weight,
        }
    }
}
impl std::cmp::Ord for Edge {
    fn cmp(&self, other: &Self) -> std::cmp::Ordering {
        (self.weight, self.src, self.dest).cmp(&(other.weight, other.src, other.dest))
    }
}
impl std::cmp::PartialOrd for Edge {
    fn partial_cmp(&self, other: &Self) -> Option<std::cmp::Ordering> {
        Some(self.cmp(other))
    }
}
type Edges = Vec<Edge>;
type Graph = Vec<Edges>;
#[allow(dead_code)]
fn graph_new(n: usize) -> Graph {
    vec![vec![]; n]
}
#[allow(dead_code)]
fn add_bi_edge(graph: &mut Graph, index: &mut usize, src: usize, dest: usize, weight: Weight) {
    graph[src].push(Edge::new(*index, src, dest, weight));
    graph[dest].push(Edge::new(*index, dest, src, weight));
    *index += 1;
}
#[allow(dead_code)]
fn add_uni_edge(graph: &mut Graph, index: &mut usize, src: usize, dest: usize, weight: Weight) {
    graph[src].push(Edge::new(*index, src, dest, weight));
    *index += 1;
}

fn main() {
    let n = read::<usize>();
    let m = read::<usize>();
    let mut g = graph_new(n);
    let mut index = 0;
    for _i in 0..m {
        let f = read::<usize>() - 1;
        let t = read::<usize>() - 1;
        add_uni_edge(&mut g, &mut index, f, t, 1);
    }
    let s = read::<usize>() - 1;
    let t = read::<usize>() - 1;
    let mut que = VecDeque::new();
    que.push_back((0, 0, s));
    let mut visited = vec![vec![false; 3]; n];
    visited[s][0] = true;
    while let Some((cost, cnt, from)) = que.pop_front() {
        if cnt == 0 && from == t {
            println!("{}", cost);
            return;
        }
        for e in g[from].iter() {
            let mut ncost = cost;
            let mut ncnt = cnt + 1;
            if ncnt == 3 {
                ncost += 1;
                ncnt = 0;
            }
            let to = e.dest;
            if visited[to][ncnt] {
                continue;
            }
            visited[to][ncnt] = true;
            que.push_back((ncost, ncnt, to));
        }
    }
    println!("{}", -1);
}