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 N = 100005;

int n, L, q;
int x[N];
int rmq[N][18];

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n;
	for (int i = 1; i <= n; ++i) cin >> x[i];
	cin >> L;
	
	int pt = 1;
	for (int i = 1; i <= n; ++i) {
		while(pt <= n && x[pt] - x[i] <= L) ++pt; --pt;
		rmq[i][0] = pt;
	}

	for (int j = 1; j <= 17; ++j) {
		for (int i = 1; i <= n; ++i) rmq[i][j] = rmq[rmq[i][j-1]][j-1];
	}

	cin >> q;
	while(q--) {
		int l, r; cin >> l >> r;
		if (l > r) swap(l, r);

		int ans = 0;
		for (int j = 17; j >= 0; --j) {
			if (rmq[l][j] < r) l = rmq[l][j], ans += (1 << j);
		}
		++ans;
		printf("%d\n", ans);
	}
}