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
from bisect import bisect
inpl = lambda: list(map(int, input().split()))

N = int(input())
X = inpl()
L = int(input())
Q = int(input())
R = [[0]*(N) for k in range(17)]

for i in range(N):
    R[0][i] = bisect(X, X[i]+L) - 1

for k in range(16):
    for i in range(N):
        R[k+1][i] = R[k][R[k][i]]


def reach(a, d):
    O = format(d, "b").zfill(17)[::-1]
    for k in range(16, -1, -1):
        if O[k] == "1":
            a = R[k][a]
    return a


def bisearch(a, b):
    a, b = a-1, b-1
    if b < a:
        a, b = b, a
    OK = b-a
    NG = 0

    while abs(OK - NG) > 1:
        mid = (OK+NG)//2
        if reach(a, mid) >= b:
            OK = mid
        else:
            NG = mid
    return OK

print(*[bisearch(*inpl()) for _ in range(Q)], sep="\n")