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 <iostream>
#include <algorithm>
#include <vector>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)

int n, l, q;
//nx[i][j] := x[j]からスタートして、2^i日後に行ける最も遠いホテルの番号(xの添字)
int nx[20][100010];
int main(void){
	cin >> n;
	vector<int> x(n);
	rep(i, n) cin >> x[i];
	cin >> l >> q;

	//ダブリング
	for (int j = 0; j < n; ++j){
		nx[0][j] = upper_bound(x.begin(), x.end(), x[j] + l) - x.begin() - 1;
	}
	for (int i = 1; i < 20; ++i){
		for (int j = 0; j < n; ++j){
			nx[i][j] = nx[i - 1][nx[i - 1][j]];
		}
	}

	for (int k = 0; k < q; ++k){
		int a, b; cin >> a >> b;
		--a; --b;
		if(b < a) swap(a, b);
		int ans = 0;
		int now = a;
		for (int i = 19; i >= 0; --i){
			if(nx[i][now] < b){
				now = nx[i][now];
				ans += 1 << i;
			}
		}
		cout << ans + 1 << endl;
	}
}