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>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}

int main() {
	int n = ri();
	int a[n];
	for (int i = 0; i < n; i++) a[i] = ri();
	int l = ri(), q = ri();
	int upper[n];
	for (int i = 0; i < n; i++) upper[i] = std::upper_bound(a, a + n, a[i] + l) - a - 1;
	std::vector<int> go[n];
	for (int i = n - 1; i >= 0; i--) {
		if (upper[i] == i) continue;
		int cur = upper[i];
		go[i].push_back(cur);
		for (int j = 0; ; j++) {
			if (j >= (int) go[cur].size() || cur == go[cur][j]) break;
			go[i].push_back(cur = go[cur][j]);
		}
	}
	for (int i = 0; i < q; i++) {
		int x = ri() - 1, y = ri() - 1;
		if (x > y) std::swap(x, y);
		int cost = 0;
		for (int i = 20; i >= 0; i--)
			if (i < (int) go[x].size() && go[x][i] < y) cost += 1 << i, x = go[x][i];
		std::cout << cost + 1 << std::endl;
	}
	return 0;
}
