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 (1.42.0) (judge verdict: Time Limit Exceeded)

use proconio::{fastout, input};
use std::io::*;
use std::str::FromStr;

use std::cmp::{max, min};

fn main() {
    input! {
        n: usize,
        m: usize,
        uv: [(usize, usize); m],
        s: usize,
        t: usize,
    }
    let mut pre_map: Vec<Vec<usize>> = vec![vec![]; n + 1];

    for i in 0..uv.len() {
        pre_map[uv[i].0].push(uv[i].1);
    }

    let mut map: Vec<Vec<usize>> = vec![vec![]; 1];
    for i in 1..=n {
        let mut ans = vec![];
        pre_dfs(i, &pre_map, 0, &mut ans);
        map.push(ans);
    }
    // println!("map: {:?}", map);
    let ans = dfs(s, &map, 0, t);
    if ans == std::i64::MAX {
        println!("-1");
        return;
    }
    println!("{}", ans);
}

fn pre_dfs(cn: usize, map: &Vec<Vec<usize>>, depth: usize, ans: &mut Vec<usize>) {
    if depth == 3 {
        ans.push(cn);
        return;
    }

    for i in 0..map[cn].len() {
        let nn = map[cn][i];
        pre_dfs(nn, map, depth + 1, ans);
    }
}

fn dfs(cn: usize, map: &Vec<Vec<usize>>, depth: usize, tn: usize) -> i64 {
    if cn == tn {
        return depth as i64;
    }

    if depth > map.len() {
        return std::i64::MAX;
    }

    let mut min_depth = std::i64::MAX;

    for i in 0..map[cn].len() {
        let nn = map[cn][i];
        min_depth = min(min_depth, dfs(nn, map, depth + 1, tn));
    }

    return min_depth;
}