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>
using namespace std;

const int MAXN = 1.1e5;
int N;
int X[MAXN];
int L;

int jump[MAXN][31];

int query(int a, int b) {
	int res = 0;
	for (int i = 30; i >= 0; i--) {
		if (jump[a][i] < b) {
			res += 1 << i;
			a = jump[a][i];
		}
	}
	assert(jump[a][0] >= b);
	return res + 1;
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);

	cin >> N;
	for (int i = 0; i < N; i++) cin >> X[i];

	cin >> L;
	for (int i = 0, j = 0; i < N; i++) {
		while (j < N && X[j] - X[i] <= L) j++;
		jump[i][0] = j-1;
	}

	for (int l = 0; l < 30; l++) {
		for (int i = 0; i < N; i++) {
			jump[i][l+1] = jump[jump[i][l]][l];
		}
	}

	int Q; cin >> Q;
	for (int q = 0; q < Q; q++) {
		int a, b; cin >> a >> b; a--, b--;
		if (a > b) swap(a, b);
		cout << query(a, b) << '\n';
	}

	return 0;
}
