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;
#define FOR(i, a, b) for(int i = (int)a; i <= (int)b; i++)

int n, arr[100050], l, q, a, b, p[100050][20];

int main() {
  	ios_base::sync_with_stdio(0);
  	cin.tie(0);
	cin >> n;
	FOR(i, 1, n) cin >> arr[i];
	cin >> l >> q;
	FOR(i, 1, n-1) p[i][0] = upper_bound(arr+1, arr+n+1, arr[i]+l) - arr - 1;
	p[n][0] = -1;
	FOR(k, 1, 17) {
		FOR(i, 1, n) {
			if (p[i][k-1] != -1) p[i][k] = p[p[i][k-1]][k-1];
			else p[i][k] = -1;
		}
	}
	while (q--) {
		cin >> a >> b;
		if (a > b) swap(a, b);
		int ans = 0;
		for(int k = 17; k >= 0; k--) {
			if (p[a][k] == -1) continue;
			if (p[a][k] <= b) {
				a = p[a][k];
				ans += (1<<k);
			}
		}
		if (a != b) ans++;
		cout << ans << "\n";
	}
}
