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
#include <bits/stdc++.h>

const int MAXN = 100010;
int xs[MAXN], D, n, Q;
int R[MAXN][20], L[MAXN][20];
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1; i <= n; ++i) std::cin >> xs[i];
	std::cin >> D >> Q;
	for (int i = 1; i <= n; ++i) {
		R[i][0] = std::upper_bound(xs + 1, xs + 1 + n, xs[i] + D) - xs - 1;
		L[i][0] = std::lower_bound(xs + 1, xs + 1 + n, xs[i] - D) - xs;
	}
	for (int i = 1; i < 20; ++i)
		for (int j = 1; j <= n; ++j)
			R[j][i] = R[R[j][i - 1]][i - 1], L[j][i] = L[L[j][i - 1]][i - 1];
	while (Q --> 0) {
		int x, y, t = 0; std::cin >> x >> y;
		if (x < y) {
		for (int i = 19; ~i; --i) if (R[x][i] < y) t += 1 << i, x = R[x][i];
		}
		else {
		for (int i = 19; ~i; --i) if (L[x][i] > y) t += 1 << i, x = L[x][i];
		}
		std::cout << t + 1 << '\n';
	}
	return 0;
}
