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: Wrong Answer)

#[allow(unused_imports)]
use {
   proconio::{fastout, input, marker::*},
   std::cmp::*,
   std::collections::*,
};

#[fastout]
fn main() {
   input! {
      n: usize,
      m: usize,
      mut uv: [(Usize1, Usize1); m+1],
   }

   if m == 0 {
      println!("-1");
      return;
   }
   let (s, t) = uv[m];
   uv.pop();

   let mut to = vec![vec![]; n];
   for &(u, v) in uv.iter() {
      to[u].push(v);
   }

   let mut to3 = vec![BTreeSet::<usize>::new(); m];
   for i in 0..n {
      dfs(i, i, 0, &to, &mut to3);
   }

   let mut d = vec![n; n];
   d[s] = 0;
   let mut q = VecDeque::new();
   q.push_back(s);
   while let Some(v) = q.pop_front() {
      for &u in to3[v].iter() {
         if d[u] > d[v] + 1 {
            d[u] = d[v] + 1;
            q.push_back(u);
         }
      }
   }
   let ans = if d[t] == n { -1 } else { d[t] as i64 };
   println!("{}", ans);
}

fn dfs(v: usize, p: usize, c: usize, to: &Vec<Vec<usize>>, to3: &mut Vec<BTreeSet<usize>>) {
   for &u in to[v].iter() {
      if to3[p].contains(&u) {
         continue;
      }
      if c == 2 {
         to3[p].insert(u);
         continue;
      }
      dfs(u, p, c + 1, to, to3);
   }
}