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.

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.

Example

Input

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


Output

4
2
1
2
import bisect

N = int(input())
x = list(map(int, input().split()))
L = int(input())

dest = []
dest_1 = [0] * N
for i in range(N):
    dest_1[i] = bisect.bisect_left(x, x[i] + L + 0.5) - 1
dest.append(dest_1)
for i in range(1, len(bin(N - 1)) - 1):
    dest_prev = dest[i - 1]
    dest_i = [0] * N
    for j in range(N):
        dest_i[j] = dest_prev[dest_prev[j]]
    dest.append(dest_i)

Q = int(input())
for _ in range(Q):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    if a > b:
        a, b = b, a
    k = len(dest) - 1
    ans = 0
    while True:
        d = dest[k][a]
        if d >= b:
            if k > 0:
                k -= 1
            else:
                ans += 1
                break
        else:
            ans += 2 ** k
            a = d
    print(ans)
