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.

-----Partial Score-----
 - 200 points will be awarded for passing the test set satisfying N \leq 10^3 and Q \leq 10^3.

-----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.

-----Sample Input-----
9
1 3 6 13 15 18 19 29 31
10
4
1 8
7 3
6 7
8 5

-----Sample Output-----
4
2
1
2

For the 1-st query, he can travel from the 1-st hotel to the 8-th hotel in 4 days, as follows:
 - Day 1: Travel from the 1-st hotel to the 2-nd hotel. The distance traveled is 2.
 - Day 2: Travel from the 2-nd hotel to the 4-th hotel. The distance traveled is 10.
 - Day 3: Travel from the 4-th hotel to the 7-th hotel. The distance traveled is 6.
 - Day 4: Travel from the 7-th hotel to the 8-th hotel. The distance traveled is 10.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'from bisect import *\n\nclass Doubling:\n    def __init__(self, A, K_max, decrement=True):\n        """\n        :param A: 写像 A:i->j を定義\n        :param K_max: K_max = 2**(k_max) まで参照する可能性がある\n        :param decrement: True = A の要素の decrement が必要\n        """\n        self.k_max = K_max.bit_length()\n        self.n = len(A)\n        self.decrement = decrement\n        self.doubling = [[-1] * self.n for _ in range(self.k_max)]\n        for i, a in enumerate(A):\n            self.doubling[0][i] = a - self.decrement\n\n        for i in range(1, self.k_max):\n            for k in range(self.n):\n                if self.doubling[i - 1][k] != -1:\n                    self.doubling[i][k] = self.doubling[i - 1][self.doubling[i - 1][k]]\n\n    def apply(self, start, K):\n        """\n        :param start: スタート地点\n        :param K: K回進む\n        :return:\n        """\n        i = start - self.decrement\n        for k in range(K.bit_length()):\n            m = 1 << k\n            if m & K:\n                i = self.doubling[k][i]\n            if i is None:\n                break\n        return i + self.decrement\n\ndef binary_search_int(ok, ng, test,a,b):\n    """\n    :param ok: solve(x) = True を必ず満たす点\n    :param ng: solve(x) = False を必ず満たす点\n    """\n    while abs(ok - ng) > 1:\n        mid = (ok + ng) // 2\n        if test(mid,a,b):\n            ok = mid\n        else:\n            ng = mid\n    return ok\n\n######################################################################################\nimport sys\ninput = sys.stdin.readline\n\nN=int(input())\nX=list(map(int, input().split()))\nL=int(input())\nQ=int(input())\n\nA=[]\nfor x in X:\n    i=bisect_left(X,x+1+L)\n    A.append(i-1)\ndoubling = Doubling(A, N+1, decrement=False)\n\ndef test(x,a,b):\n    return X[doubling.apply(a,x)]<X[b]\n\nfor _ in range(Q):\n    a,b=map(int, input().split())\n    a,b=a-1,b-1\n    if a>b: a,b=b,a\n    print(binary_search_int(0,N,test,a,b)+1)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 16661