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::{ BinaryHeap, VecDeque };

struct Graph {
    edges: Vec<Vec<(usize, i64)>>, // adjacent list
    n: usize
}

impl Graph {
    fn new(n: usize) -> Self {
        Graph {
            edges: vec![Vec::new(); n],
            n: n
        }
    }

    fn add_costs(&mut self, from: usize, to: usize, cost: i64, directed: bool) {
        if directed {
            self.edges[from].push((to, cost));
        } else {
            self.edges[from].push((to, cost));
            self.edges[to].push((from, cost));
        }

    }

    fn dijkstra_search(&self, start: usize) -> Vec<i64> {
        const INF: i64 = 1 << 61;
        let mut dist: Vec<i64> = vec![INF; self.n];
        let mut visited: Vec<bool> = vec![false; self.n];
        dist[start] = 0;
        let mut que: BinaryHeap<(i64, usize)> = BinaryHeap::new();
        que.push((0, start));
        while let Some((_, v)) = que.pop() {
            if visited[v] {
                continue
            }
            visited[v] = true;
            for edge in &self.edges[v] {
                let (u, cost) = *edge;
                if visited[u] {
                    continue
                }
                if dist[u] > dist[v] + cost {
                    dist[u] = dist[v] + cost;
                    que.push(( -dist[u], u));
                }
            }
        }
        return dist
    }
}

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    input! {
    n: usize,
    m: usize,
    uv: [(usize, usize); m],
    s: usize,
    t: usize,
    }
    let mut graph = Graph::new(3*n+1);
    for i in 0..m {
        let u = uv[i].0;
        let v = uv[i].1;
        graph.add_costs(u, v+n, 1, true);
        graph.add_costs(u+n, v+2*n, 1, true);
        graph.add_costs(u+2*n, v, 1, true);
    }

    let result = graph.dijkstra_search(s);
    const INF: i64 = 1 << 61;
    if result[t] == INF {
        println!("{}", -1);
    } else {
        println!("{}", result[t]/3)
    }
}