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
def I(): return int(sys.stdin.readline().rstrip())
def MI(): return map(int,sys.stdin.readline().rstrip().split())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))  #空白あり


N = I()
x = LI()
L = I()

from bisect import bisect_right

arrive = [[0]*N for _ in range(30)]  # arrive[k][i] = i番目(0-indexed)のホテルから2**k日で到達できる最も遠いホテル
for k in range(30):
    if k == 0:
        for i in range(N):
            arrive[0][i] = bisect_right(x,x[i]+L)-1
    else:
        for i in range(N):
            arrive[k][i] = arrive[k-1][arrive[k-1][i]]


def query(a,b):  # (a-1)番目のホテルから(b-1)番目の駅まで移動するのにかかる日数
    a -= 1
    b -= 1
    ans = 1
    for k in range(29,-1,-1):
        if arrive[k][a] >= b:
            continue
        else:
            a = arrive[k][a]
            ans += 1 << k
    print(ans)


Q = I()
for i in range(Q):
    a,b = MI()
    if a > b:
        a,b = b,a
    query(a,b)
