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
input = sys.stdin.readline
from bisect import bisect
N=int(input())
X=[int(i) for i in input().split()]
L=int(input())
Q=int(input())
#T=[[int(i) for i in input().split()] for i in range(Q)]
T=[]
for i in range(Q):
    a,b=map(int,input().split())
    if a>b:
        a,b=b,a
    T.append((a-1,b-1))
F=[0]*N
for i in range(N):
    F[i]= bisect(X, X[i] + L) -1
#print(F)
dp=[ [N-1] * (N) for i in range(18)]

for i in range(N):
    #dp[0][i]=i
    dp[0][i]=F[i]
for c in range(1,18):
    for i in range(N):
        dp[c][i] = dp[c-1][ dp[c-1][i] ]
#print(dp)

def f(a,b):
    num=0
    t = a
    while True:
        for i in range(18):
            if dp[i][t]>=b:
                s=i
                break
        if s==0:
            num+=1
            break
        num += pow(2, s - 1)
        t = dp[s-1][t]
    print(num)
for a,b in T:
    f(a,b)

