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 = 100000;
const int maxh = 17;

int N;
int x[maxN];
int L;
int nxt[maxN];
int f[maxN][maxh];

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);

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

	for (int i = 0; i < N; i++)
		nxt[i] = upper_bound(x + i + 1, x + N, x[i] + L) - 1 - x;

	for (int h = 0; h < maxh; h++) {
		for (int i = 0; i < N; i++)
			f[i][h] = h == 0 ? nxt[i] : f[f[i][h - 1]][h - 1];
	}

	int Q;
	cin >> Q;
	for (int QN = 0; QN < Q; QN++) {
		int a, b;
		cin >> a >> b;
		a--, b--;
		if (a > b)
			swap(a, b);

		int ans = 0;
		for (int h = maxh - 1; h >= 0; h--) {
			if (f[a][h] < b) {
				a = f[a][h];
				ans += 1 << h;
			}
		}
		ans++;

		cout << ans << '\n';
	}

	return 0;
}
