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
from bisect import bisect_left, bisect_right
import math
import sys
input = sys.stdin.readline


def main():
    N = int(input())
    X = list(map(int, input().split()))
    L = int(input())
    ln = int(math.log(N, 2))
    dpl = [[0] * N for _ in range(ln+1)]
    dpr = [[0] * N for _ in range(ln+1)]
    l = 0
    r = 0
    for i in range(N):
        x = X[i]
        while X[l] + L < x:
            l += 1
        while r < N-1 and x + L >= X[r+1]:
            r += 1
        dpl[0][i] = l
        dpr[0][i] = r
    for k in range(1, ln+1):
        for i in range(N):
            dpl[k][i] = dpl[k-1][dpl[k-1][i]]
            dpr[k][i] = dpr[k-1][dpr[k-1][i]]
    Q = int(input())
    for _ in range(Q):
        a, b = map(int, input().split())
        a -= 1
        b -= 1
        ans = 0
        if a < b:
            for k in range(ln+1)[::-1]:
                if dpr[k][a] < b:
                    a = dpr[k][a]
                    ans += 2**k
            if a < b:
                ans += 1
        else:
            # a > b
            for k in range(ln+1)[::-1]:
                if dpl[k][a] > b:
                    a = dpl[k][a]
                    ans += 2**k
            if a > b:
                ans += 1
        print(ans)


if __name__ == "__main__":
    main()
