Problem:
Score : 700 points

Problem Statement

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.

Partial Score

200 points will be awarded for passing the test set satisfying N \leq 10^3 and Q \leq 10^3.

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.

Sample Input 1

9
1 3 6 13 15 18 19 29 31
10
4
1 8
7 3
6 7
8 5

Sample Output 1

4
2
1
2

For the 1-st query, he can travel from the 1-st hotel to the 8-th hotel in 4 days, as follows:

Day 1: Travel from the 1-st hotel to the 2-nd hotel. The distance traveled is 2.

Day 2: Travel from the 2-nd hotel to the 4-th hotel. The distance traveled is 10.

Day 3: Travel from the 4-th hotel to the 7-th hotel. The distance traveled is 6.

Day 4: Travel from the 7-th hotel to the 8-th hotel. The distance traveled is 10.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
 
#define min(a,b) (((a)<(b))?(a):(b))
#define max(a,b) (((a)>(b))?(a):(b))

int main() {
	long N, L, Q, B;
	long x[100001];
	long i, j, k;
	long a, b, tmp;
	long move1day[100001];
	long moveBday[100001] = {0};
	long right;
	long ans[100001] = {0};
	
	scanf("%ld", &N);
	B = sqrt(N);
	for(i=1;i<=N;i++) scanf("%ld", &x[i]);
	scanf("%ld", &L);
	
	/* 各ホテルから1日でどこまでいけるか計算 O(N) */
	right = 2;
	for(i=1;i<=N;i++) {
		for(j=right;j<=N;j++) {
			if(x[j] - x[i] > L) {
				j--;
				break;
			}
		}
		move1day[i] = j;
		right = j;
	}
	
	/* 各ホテルから√N日でどこまでいけるか計算 O(N√N) */
	for(i=1;i<=N;i++) {
		k = i;
		for(j=0;j<B;j++) {
			if(k >= N) break;
			k = move1day[k];
		}
		if(j == B) moveBday[i] = k;
	}
	
	scanf("%ld", &Q);
	for(i=1;i<=Q;i++) {
		scanf("%ld %ld", &a, &b);
		
		if(a > b) { tmp = a; a = b; b = tmp; }
		
		/* √N日ずつ移動する */
		j = a;
		while(moveBday[j] > 0 && moveBday[j] < b) {
			j = moveBday[j];
			ans[i] += B;
		}
		
		/* 1日ずつ移動する */
		while(j < b) {
			j = move1day[j];
			ans[i]++;
		}
	}
	
	for(i=1;i<=Q;i++) printf("%ld\n", ans[i]);
	
	return 0;
}