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<stdint.h>
#include<inttypes.h>

typedef int32_t i32;

#define POS(i, j) ((i) * (bit + 1) + (j))

void run (void) {
  i32 n;
  scanf ("%" SCNi32, &n);
  i32 *x = (i32 *) calloc (n, sizeof (i32));
  for (i32 i = 0; i < n; ++i) {
    scanf ("%" SCNi32, x + i);
  }
  i32 l;
  scanf ("%" SCNi32, &l);
  i32 bit = 0;
  while ((1 << bit) < n) bit++;
  i32 *dp = (i32 *) calloc (n * (bit + 1), sizeof (i32));
  for (i32 i = 0, j = 0; i < n; ++i) {
    while (j < n && x[j] - x[i] <= l) j++;
    dp[POS(i, 0)] = j - 1;
  }
  for (i32 k = 1; k <= bit; ++k) {
    for (i32 i = 0; i < n; ++i) {
      i32 a = dp[POS(i, k - 1)];
      dp[POS(i, k)] = dp[POS(a, k - 1)];
    }
  }
  i32 q;
  scanf ("%" SCNi32, &q);
  while (q--) {
    i32 a, b;
    scanf ("%" SCNi32 "%" SCNi32, &a, &b);
    a--;
    b--;
    if (a > b) {
      i32 s = b;
      b = a;
      a = s;
    }
    i32 ans = 0;
    i32 j = bit;
    while (a < b) {
      while (j > 0 && dp[POS(a, j)] > b) --j;
      ans += 1 << j;
      a = dp[POS(a, j)];
    }
    printf ("%" PRIi32 "\n", ans);
  }
}

int main (void) {
  run ();
  return 0;
}