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
# seishin.py
N = int(input())
*X, = map(int, input().split())
L = int(input())

K = 20

D = [[N]*(K+1) for i in range(N)]
j = N-1
for i in range(N-1, -1, -1):
    while j > 0 and X[j] - X[i] > L:
        j -= 1
    D[i][0] = j
for k in range(K):
    for i in range(N):
        if D[i][k] != N:
            D[i][k+1] = D[D[i][k]][k]
def solve(a, b):
    if not a < b:
        a, b = b, a
    res = 1
    for k in range(K, -1, -1):
        if D[a][k] < b:
            a = D[a][k]
            res += 2**k
    return res
Q = int(input())
ans = []
for i in range(Q):
    a, b = map(int, input().split())
    ans.append(solve(a-1, b-1))
print(*ans, sep='\n')
