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;

const int N = 1e5;
const int lgn = 17;

int n,q,L;
int li[N+5];

int sparse[lgn+1][N+5];

void make_sparse(){
	for(int i=1;i<=lgn;i++){
		for(int j=1;j<=N;j++){
			sparse[i][j] = sparse[i-1][sparse[i-1][j]];
		}
	}
	return;
}

int find_day(int x, int y){
	int ret = 0;
	for(int i=lgn;i>=0;i--){
		if(sparse[i][x]<y){
			ret+=(1<<i);
			x = sparse[i][x];
		}
	}
	return ret+1;
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&li[i]);
	}
	scanf("%d",&L);
	scanf("%d",&q);
	int r = 1;
	for(int i=1;i<=n;i++){
		while(r <= n && li[r] - li[i] <= L) r++;
		sparse[0][i] = r-1;
	}
	make_sparse();
	for(int i=1;i<=q;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		if(u>v) swap(u,v);
		int ans = find_day(u,v);
		printf("%d\n",ans);
	}
	return 0;
}