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 *

def judge(a, b, x):
    now = a
    c = 0
    
    while x>0:
        if x&1:
            now = dp[c][now]
        
        x >>= 1
        c += 1
    
    return now>=b
    
def binary_search(a, b):
    l, r = 0, N-1
    
    while l<=r:
        m = (l+r)//2
        
        if judge(a, b, m):
            r = m-1
        else:
            l = m+1
    
    return l

N = int(input())
x = list(map(int, input().split()))
L = int(input())
log_size = 22
dp = [[0]*N for _ in range(log_size)]

for i in range(N):
    dp[0][i] = bisect_right(x, x[i]+L)-1
    
for i in range(1, log_size):
    for j in range(N):
        dp[i][j] = dp[i-1][dp[i-1][j]]

Q = int(input())

for _ in range(Q):
    a, b = map(int, input().split())
    a, b = min(a, b)-1, max(a, b)-1
    print(binary_search(a, b))