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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int a[100010], b[100010], x[100010], r[100010][20];

int intcmp(const void *a, const void *b) {
	int c = *(int *)a-*(int *)b;
	if (c>0) return 1;
	else if (!c) return 0;
	else return -1;
}

int main() {
	int D, from, i, j, k, key, L, N, *ptr, Q, to;
	scanf("%d", &N);
	for (i=0;i<N;i++) scanf("%d", &x[i]);
	scanf("%d%d", &L, &Q);
	for (i=0;i<Q;i++) scanf("%d%d", &a[i], &b[i]);

	for (i=0;i<N-1;i++) {
		j = 0;
		key = x[i]+L;
		ptr = bsearch(&key, x, N, sizeof(int), intcmp);
		while (ptr == NULL) { --key; ptr = bsearch(&key, x, N, sizeof(int), intcmp); }
		r[i][0] = (ptr-x)+1;
	}
	r[N-1][0] = N;
	for (k=1;k<20;k++) {
		for (i=0;i<N-1;i++) {
			r[i][k] = r[r[i][k-1]-1][k-1];
		}
		r[N-1][k] = N;
	}

	for (i=0;i<Q;i++) {
		D = 0;
		if (a[i]>b[i]) {
			from = b[i]; to = a[i];
			while ((ptr = bsearch(&to, r[from-1], 20, sizeof(int), intcmp)) == NULL) { --to; }
			k = (ptr-r[from-1])+1;
			j = 1;
			while (--k) j *= 2;
			D += j;
			while (to != a[i]) {
				from = to; to = a[i];
				while ((ptr = bsearch(&to, r[from-1], 20, sizeof(int), intcmp)) == NULL) { --to; }
				k = (ptr-r[from-1])+1;
				j = 1;
				while (--k) j *= 2;
				D += j;
			}
		} else {
			from = a[i]; to = b[i];
			while ((ptr = bsearch(&to, r[from-1], 20, sizeof(int), intcmp)) == NULL) { --to; }
			k = (ptr-r[from-1])+1;
			j = 1;
			while (--k) j *= 2;
			D += j;
			while (to != b[i]) {
				from = to; to = b[i];
				while ((ptr = bsearch(&to, r[from-1], 20, sizeof(int), intcmp)) == NULL) { --to; }
				k = (ptr-r[from-1])+1;
				j = 1;
				while (--k) j *= 2;
				D += j;
			}
		}
		printf("%d\n", D);
	}
	return 0;
}