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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>) {
    val M = readLine()!!.toInt()
    var N = 1
    while (N < M) N *= 2
    val x = readLine()!!.split(" ").map{it.toLong()}
    val L = readLine()!!.toLong()
    val Q = readLine()!!.toInt()
    val a = Array(Q) { 0 }
    val b = Array(Q) { 0 }

    for (i in a.indices) {
        val (A, B) = readLine()!!.split(" ").map{it.toInt()}
        a[i] = Math.min(A, B) - 1
        b[i] = Math.max(A, B) - 1
    }

    val mp = TreeMap<Long, Int>()
    x.forEachIndexed { index, l -> mp[l] = index }

    val doubling = Array(N) { Array(20) { N - 1 } }

    for (i in x.indices) {
        doubling[i][0] = mp.floorEntry(x[i] + L)?.value ?: mp.lastEntry().value
    }

    for (k in 1 until 20) {
        for (i in doubling.indices) {
            doubling[i][k] = doubling[doubling[i][k - 1]][k - 1]
        }
    }

    fun findNext(s: Int, t: Int): Pair<Int, Int> {
        var k = doubling[s].indexOfFirst { it >= t }
        if (doubling[s][k] > t && k > 0) k--

        return Pair(doubling[s][k], 1 shl k)
    }

    for (l in a.indices) {
        var res = 0

        var v = a[l]
        while (v < b[l]) {
            val p = findNext(v, b[l])
            v = p.first
            res += p.second
        }
        println(res)
    }
}