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 Python (3.4.3) (judge verdict: Accepted)

import sys
sys.setrecursionlimit(10**7)
INF = 10 ** 18
MOD = 10 ** 9 + 7
from functools import partial, reduce
from operator import mul
prod = partial(reduce, mul)
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x) - 1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def II(): return int(sys.stdin.readline())
def SI(): return input()

from bisect import bisect

def main():
    N = II()
    X = LI()  # coordinate of hotels
    X.append(INF)  # banpei
    L = II()  # max move distance/day
    Q = II()
    AB = []  # two hotel (a,b)
    for _ in range(Q):
        AB.append(LI_())
    Unreachs = [[] for _ in range(N)]
    # 1day
    for i, x in enumerate(X[:-1]):
        u = bisect(X, x + L)
        Unreachs[i].append(u)
    # day 2, 4, 8, ...
    updated = True
    while updated:
        updated = False
        for i in range(N):
            u = Unreachs[Unreachs[i][-1] - 1][-1]
            updated = updated or (u != Unreachs[i][-1])
            Unreachs[i].append(u)
    # solve
    for a, b in AB:
        if a > b:
            a, b = b, a
        ans = 0
        while True:
            k = bisect(Unreachs[a], b) - 1
            if k < 0:
                ans += 1
                break
            ans += 2 ** k
            a = Unreachs[a][k] - 1
        print(ans)
    return 0

main()