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<stdio.h>
#include<algorithm>
using namespace std;
int n,l,a[111111];
int nex[111111][22];
int calc(int x,int y){
	if(nex[x][0] >= y)return 1;
	int p = 0;
	while(nex[x][p] < y)p++;
	--p;
	return (1<<p) + calc(nex[x][p],y);
}
int main(){
	scanf("%d",&n);
	for(int i=1; i<=n; i++)
		scanf("%d",&a[i]);
	scanf("%d", &l);
	for(int i=n; i>=1; i--){
		if(i==n){
			for(int j=0; j<=17; j++)
				nex[i][j] = n;
		}else{
			nex[i][0] = upper_bound(a+1,a+n+1,a[i]+l)-a-1;
			for(int j=1; j<=17; j++){
				nex[i][j] = nex[nex[i][j-1]][j-1];
			}
		}
	}
	int q;
	for(scanf("%d",&q); q--;){
		int x,y;
		scanf("%d%d", &x, &y);
		if(x>y)swap(x,y);
		printf("%d\n",calc(x,y));
	}
	return 0;
}