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
import sys
import math
import itertools
sys.setrecursionlimit(10000)

INF = float('inf')

# input macro
def i():
    return int(raw_input())
def ii():
    return map(int,raw_input().split(" "))
def s():
    return raw_input()
def ss():
    return raw_input().split(" ")
def slist():
    return list(raw_input())
def join(s):
    return ''.join(s)

###########

n=i()
x=ii()
l=i()
q=i()
a,b=[0]*q,[0]*q
for j in range(q):
    aa,bb=ii()
    a[j],b[j]=min(aa,bb)-1,max(aa,bb)-1
table=[[0 for i in range(n)] for j in range(32)]


for k in range(n):
    table[0][k]=bisect.bisect_right(x,x[k]+l)-1
for j in range(1,32):
    for k in range(n):
        table[j][k]=table[j-1][table[j-1][k]]
for j in range(q):
    day=0
    cur=a[j]
    for k in reversed(range(32)):
        if table[k][cur]<b[j]:
            day+=2**k
            cur=table[k][cur]
    print day+1
