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;

int n, l, q;
long long x[111111];
int f[111111][22];

int main(){
//	freopen("input.inp", "r", stdin);
	scanf("%d",&n);
	for(int i = 1; i <= n; ++i) scanf("%lld",&x[i]);
	scanf("%d%d",&l,&q);
	for(int i = 1; i <= n; ++i){
		int ll = i + 1, rr = n;
		while (ll <= rr){
			int md = (ll + rr) >> 1;
			if (x[md] - x[i] <= l) ll = md + 1;
			else rr = md - 1;
		}
		f[i][0] = rr;
	}
	for(int j = 1; j < 17; ++j){
		for(int i = 1; i <= n; ++i){
			f[i][j] = f[f[i][j - 1]][j - 1];
		}
	}
	while (q--){
		int a, b;
		scanf("%d%d",&a,&b);
		if (a > b) swap(a, b);
		int an = 0;
		for(int i = 16; i >= 0; --i){
			if (f[a][i] < b) a = f[a][i], an += (1 << i);
		}
		printf("%d\n",an + 1);
	}
	return 0;
}