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)

// https://atcoder.jp/contests/arc060/tasks/arc060_c
//
#![allow(unused_imports)]
use std::io::*;
use std::io::Write;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;

// Input macros.
// Original by tanakh: https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

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

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

#[allow(unused_macros)]
macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

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

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

    ($iter:expr, switch) => {
        {
            let ty = read_value!($iter, i32);
            if ty == 1 {
                vec![ty, read_value!($iter, i32), read_value!($iter, i32)]
            } else if ty == 2 {
                vec![ty, read_value!($iter, i32)]
            } else {
                vec![ty, read_value!($iter, i32)]
            }
        }
    };

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

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

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

#[allow(unused_macros)]
macro_rules! read_line {
    ($t:tt) => {
        {
            let mut s = String::new();
            std::io::stdin().read_line(&mut s).unwrap();
            s.trim_right().parse::<$t>().unwrap()
        }
    }
}

#[allow(unused_macros)]
macro_rules! dvec {
    ($t:expr ; $len:expr) => {
        vec![$t; $len]
    };

    ($t:expr ; $len:expr, $($rest:expr),*) => {
        vec![dvec!($t; $($rest),*); $len]
    };
}

#[allow(unused_macros)]
macro_rules! ifv {
    ($t:expr, $a:expr, $b: expr) => {
        if $t { $a } else { $b }
    }
}

#[allow(unused_macros)]
macro_rules! fill {
    ($t:expr, $v:expr) => {
        for i in 0..$t.len() {
            $t[i] = $v;
        }
    };
}

#[allow(unused_macros)]
macro_rules! join {
    ($t:expr, $glue:expr) => {
        $t.into_iter().map(|w| w.to_string()).collect::<Vec<_>>().join($glue)
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        println!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

// ===

fn build(limit: i64, x: &Vec<i64>, dbl: usize) -> Vec<Vec<usize>> {
    let n = x.len();
    let mut wnext = dvec!(0; dbl, n);
    let mut to = 0;
    for i in 0..n {
        to = max(to, i);
        while to+1 < n && x[to+1] - x[i] <= limit {
            to += 1;
        }
        wnext[0][i] = to;
    }
    for k in 1..dbl {
        for i in 0..n {
            wnext[k][i] = wnext[k-1][wnext[k-1][i]];
        }
    }
    wnext
}

fn main() {
    input! {
        n: usize,
        mut x: [i64; n],
        limit: i64,
        q: usize,
        queries: [(usize1, usize1); q]
    };

    let dbl = 20;
    let wright = build(limit, &x, dbl);
    x.reverse();
    for i in 0..n {
        x[i] *= -1;
    }
    let wleft_rev = build(limit, &x, dbl);
    let mut wleft = dvec!(0; dbl, n);
    for k in 0..dbl {
        for i in 0..n {
            wleft[k][i] = n-1-wleft_rev[k][n-1-i];
        }
    }

    let mut ans = vec![0; n];
    for (a, b) in queries {
        let mut now = a;
        let mut cost = 0;
        if a < b {
            for k in (0..dbl).rev() {
                if wright[k][now] < b {
                    cost += 1<<k;
                    now = wright[k][now];
                }
            }
        } else if b < a {
            for k in (0..dbl).rev() {
                if wleft[k][now] > b {
                    cost += 1<<k;
                    now = wleft[k][now];
                }
            }
        }
        println!("{}", cost+1);
    }
}