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 sys
sys.setrecursionlimit(10 ** 6)  # 変更
from bisect import bisect_left,bisect

n = int(input())
x = list(map(int,input().split()))
L = int(input())
t = n.bit_length()
prv = [n-1] * n
for i in range(n):
    prv[i] = bisect(x,x[i]+L) - 1
kprv = [prv]
S = prv
for i in range(t-1):
    kar = [n-1] * n
    for j in range(n):
        kar[j] = S[S[j]]
    kprv.append(kar)
    S = kar

q = int(input())
for _ in range(q):
    a , b = map(lambda x:int(x)-1,input().split())
    if a > b:
        a , b = b , a
    lef = 0
    rig = n-1
    while rig - lef > 1:
        mid = (rig+lef)//2
        p = mid.bit_length()
        now = a
        for i in range(p):
            if (mid>>i) & 1 == 1:
                now = kprv[i][now]
        if now >= b:
            rig = mid
        elif now < b:
            lef = mid
    print(rig)