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>
#define ll long long
using namespace std;

int n, l, q, ai, bi; 
int a[1000010];
int f[1000010][40];

void prepare() {
	for(int i = 1; i <= n; i++){
		int id = upper_bound(a + 1,a + 1 + n, a[i] + l) - a - 1;  
		if(a[i] + l > a[n])	f[i][0] = n;
		else f[i][0] = id;
	} 
	for(int i = 1; i <= 32; i++)
		for(int j = 1; j <= n; j++)
			f[j][i] = f[f[j][i - 1]][i - 1];
}

int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; i++) 
		scanf("%d", &a[i]);
	scanf("%d%d", &l, &q);
	prepare();
	while(q--) {
		scanf("%d%d", &ai, &bi);
		if(ai > bi) swap(ai, bi);
		int ans = 0;
		for(int j = 30; j >= 0; j --)  
			if(f[ai][j] < bi) { ans += (1 << j); ai = f[ai][j];}
		printf("%d\n", ans + 1);
	}
	return 0;
} 