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
n = int(input())
x = list(map(int, input().split()))
l = int(input())
q = int(input())
ab=[list(map(int,input().split())) for i in range(q)]

keta=len(format(n+1,"b"))
nxt = [[0 for _ in range(n)] for _ in range(keta)]
# nxt[k][node] == farthest node-num from node in 2**k days

import bisect
for i in range(n):
    num = x[i]+l
    ind = bisect.bisect_right(x,num)
    nxt[0][i] = ind-1

for k in range(1,keta):
    for i in range(n):
        nxt[k][i]=nxt[k-1][nxt[k-1][i]]

def solve(s,g,day):
    bi=format(day,"b")
    keta=len(bi)-1

    for b in bi:
        if b=="1":
            s=nxt[keta][s]
        keta-=1

    return s>=g
for a,b in ab:
    s=min(a,b)
    g=max(a,b)
    l=0
    r=n
    while r-l>1:
        mid=(l+r)//2
        if solve(s-1,g-1,mid):
            r=mid
        else:
            l=mid

    print(r)