Problem:
Score : 700 points

Problem Statement

N hotels are located on a straight line. The coordinate of the i-th hotel (1 \leq i \leq N) is x_i.

Tak the traveler has the following two personal principles:

He never travels a distance of more than L in a single day.

He never sleeps in the open. That is, he must stay at a hotel at the end of a day.

You are given Q queries. The j-th (1 \leq j \leq Q) query is described by two distinct integers a_j and b_j.
For each query, find the minimum number of days that Tak needs to travel from the a_j-th hotel to the b_j-th hotel following his principles.
It is guaranteed that he can always travel from the a_j-th hotel to the b_j-th hotel, in any given input.

Constraints

2 \leq N \leq 10^5

1 \leq L \leq 10^9

1 \leq Q \leq 10^5

1 \leq x_i < x_2 < ... < x_N \leq 10^9

x_{i+1} - x_i \leq L

1 \leq a_j,b_j \leq N

a_j \neq b_j

N,\,L,\,Q,\,x_i,\,a_j,\,b_j are integers.

Partial Score

200 points will be awarded for passing the test set satisfying N \leq 10^3 and Q \leq 10^3.

Input

The input is given from Standard Input in the following format:

N
x_1 x_2 ... x_N
L
Q
a_1 b_1
a_2 b_2
:
a_Q b_Q

Output

Print Q lines.
The j-th line (1 \leq j \leq Q) should contain the minimum number of days that Tak needs to travel from the a_j-th hotel to the b_j-th hotel.

Sample Input 1

9
1 3 6 13 15 18 19 29 31
10
4
1 8
7 3
6 7
8 5

Sample Output 1

4
2
1
2

For the 1-st query, he can travel from the 1-st hotel to the 8-th hotel in 4 days, as follows:

Day 1: Travel from the 1-st hotel to the 2-nd hotel. The distance traveled is 2.

Day 2: Travel from the 2-nd hotel to the 4-th hotel. The distance traveled is 10.

Day 3: Travel from the 4-th hotel to the 7-th hotel. The distance traveled is 6.

Day 4: Travel from the 7-th hotel to the 8-th hotel. The distance traveled is 10.

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::io::BufRead;

fn main() {
    let stdin = std::io::stdin();
    let n = stdin.lock().gets().parse::<usize>().unwrap();
    let hotels = stdin.lock().gets().split(' ').take(n)
        .map(|x| x.parse().unwrap()).collect::<Vec<u32>>();
    let ell = stdin.lock().gets().parse::<u32>().unwrap();
    let q = stdin.lock().gets().parse::<usize>().unwrap();
    let mut queries = Vec::with_capacity(q);
    for line in stdin.lock().lines().take(q).map(|x| x.unwrap()) {
        let (a, b): (usize, usize) = line.split(' ').values();
        queries.push(if a < b { (a - 1, b - 1) } else { (b - 1, a - 1) });
    }

    let mut next = vec![0; n];
    for i in 0..n {
        match hotels[i + 1..].binary_search(&(hotels[i] + ell + 1)) {
            Ok(pos) | Err(pos) => next[i] = i + 1 + pos - 1,
        }
    }
    // Exponential jump table.
    let mut jt = vec![Vec::new(); n];
    for i in (0..n).rev() {
        let mut lv = 0;
        let mut p = next[i];
        jt[i].push(p);
        while p < n - 1 && lv < jt[p].len() {
            p = jt[p][lv];
            jt[i].push(p);
            lv += 1;
        }
    }
    for (a, b) in queries {
        let mut x = a;
        let mut days = 0;
        while x < b {
            match jt[x].iter().enumerate().rev().find(|&(_, &to)| to <= b) {
                Some((i, &to)) => {
                    x = to;
                    days += 1 << i;
                },
                None => {
                    days += 1;	// can go beyond "b" in one day.
                    break;
                },
            }
        }
        println!("{}", days);
    }
}

trait Gets { fn gets(&mut self) -> String; }
impl<T: std::io::BufRead> Gets for T {
    fn gets(&mut self) -> String { self.lines().next().unwrap().unwrap() } }

trait Values<T> { fn values(&mut self) -> T; }
impl<'a, I, T1, T2> Values<(T1, T2)> for I
    where I: Iterator<Item=&'a str>,
          T1: std::str::FromStr, T1::Err: std::fmt::Debug,
          T2: std::str::FromStr, T2::Err: std::fmt::Debug {
    fn values(&mut self) -> (T1, T2) {
        (self.next().unwrap().parse().unwrap(),
         self.next().unwrap().parse().unwrap())
    }
}