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
#!/usr/bin/env python3
import bisect


def dist(tree, tree_i, tree_j, parent, a, b):
    i = tree_i[b]
    if tree_i[a] == i:
        return tree_j[b] - tree_j[a]
    else:
        branch = tree[i]
        branch_root = branch[0]
        if branch_root < a:
            return tree_j[b] - bisect.bisect_left(branch, a) + 1
        else:
            p = parent[branch_root]
            if p < a:
                return tree_j[b] + 1
            else:
                return tree_j[b] + 1 + dist(tree, tree_i, tree_j, parent, a, p)


def solve(n, x, l, q, qry):

    g = [[] for _ in range(n)]
    parent = [-1] * n
    weight = [1] * n

    for v in range(1, n):
        p = bisect.bisect_left(x, x[v] - l)
        g[p].append(v)
        parent[v] = p

    for v in range(n - 1, -1, -1):
        for w in g[v]:
            weight[v] += weight[w]

    tree = [[0]]
    tree_i = [-1] * n
    tree_j = [0] * n
    tree_i[0] = 0
    for v in range(n):
        mw = 0
        c = -1
        for w in g[v]:
            if mw < weight[w]:
                mw = weight[w]
                c = w
        if 0 <= c:
            i = tree_i[v]
            tree[i].append(c)
            tree_i[c] = i
            tree_j[c] = len(tree[i]) - 1
            for w in g[v]:
                if w != c:
                    tree.append([w])
                    tree_i[w] = len(tree) - 1

    for tpl in qry:
        a, b = tpl
        if b < a:
            a, b = b, a
        print(dist(tree, tree_i, tree_j, parent, a, b))


def main():
    n = input()
    n = int(n)
    x = list(map(int, input().split()))
    l = input()
    l = int(l)
    q = input()
    q = int(q)
    qry = []
    for _ in range(q):
        a, b = input().split()
        a = int(a) - 1
        b = int(b) - 1
        qry.append((a, b))

    solve(n, x, l, q, qry)


if __name__ == '__main__':
    main()

