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: Time Limit Exceeded)

from bisect import bisect, bisect_left

n = int(input())
x = list(map(int, input().split()))
l = int(input())

def search_1(start):
	return bisect(x, x[start] + l) - 1

route = [[] for _ in range(n)]

i = 0 # 親の位置
for i in range(n - 1):
	if route[i] == []:
		j = search_1(i) # カーソルの位置
		num = 1 # 何日目の場所か
		route[i].append(j)
		if route[j] == []:
			route[j] = (i, num)
		else: # 既に他の親が通っている
			route[i] += route[route[j][0]][route[j][1]:]
			continue
		while j < n - 1:
			j = search_1(j)
			num += 1
			route[i].append(j)
			if route[j] == []:
				route[j] = (i, num)
			else: # 既に他の親が通っている
				route[i] += route[route[j][0]][route[j][1]:]
				break
	else:
		continue

# print(route)

def search_2(start, goal):
	if type(route[start]) == type([]):
		return bisect_left(route[start], goal) + 1
	else:
		return bisect_left(route[route[start][0]][route[start][1]:], goal) + 1



q = int(input())
for _ in range(q):
	a, b = map(int, input().split())
	if a > b:
		a, b = b, a
	print(search_2(a - 1, b - 1))