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 C getchar()
inline int read() {
	int num = 0;
	char c = C;
	for(;c<'0' || c>'9';c=C);
	for(;c>='0' && c<='9';c=C) num = (num << 1) + (num << 3) + c - 48;
	return num;
}

int n,k,q;
int a[110000];
int f[110000][25];

int main() {
	
	n = read();
	for(int i=1;i<=n;++i) a[i] = read();
	k = read(),q = read();
	a[n + 1] = 2e9;
	int now = 1;
	
	for(int i=1;i<=n;++i) {
		while (a[now] - a[i] <= k) ++ now;
		f[i][0] = now - 1;
	}
	
	for(int i=1;1<<i<=n;++i) {
		for(int j=n;j>=1;--j) { 
			f[j][i] = f[f[j][i - 1]][i - 1];
		}
	}
	
	for(int i=1;i<=q;++i) {
		int x = read(),y = read();
		if (x > y) swap(x , y);
		int j = 20,ans = 0;
		for(;1 << j > n;--j);
		
		for(;j>=0;--j) while (f[x][j] < y) x = f[x][j],ans += 1 << j;
		printf("%d\n",ans + 1);
	}
	
	return 0;
} 