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<math.h>

typedef long long int int64;

#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define ABS(a) ((a)>(0)?(a):-(a))

#define M 18

#define POS(i,j) ((i)*M+(j))

void run(void){
  int n;
  scanf("%d",&n);
  int *x=(int *)malloc(sizeof(int)*n);
  int i;
  for(i=0;i<n;i++) scanf("%d",x+i);
  int l;
  scanf("%d",&l);
  int *dp=(int *)malloc(sizeof(int)*n*M);
  int r=0;
  for(i=0;i<n;i++){
    while(r<n && x[r]-x[i]<=l) r++;
    dp[POS(i,0)]=r-1;
  }
  for(i=0;dp[POS(0,i)]<n-1;i++){
    int j;
    for(j=0;j<n;j++){
      dp[POS(j,i+1)]=dp[POS(dp[POS(j,i)],i)];
    }
  }
  int q;
  scanf("%d",&q);
  while(q--){
    int a,b;
    scanf("%d%d",&a,&b);
    a--;
    b--;
    if(a>b){
      int c=a;
      a=b;
      b=c;
    }
    int ans=0;
    int now=a;
    while(now<b){
      int i=0;
      while(dp[POS(now,i)]<b) i++;
      if(i==0){
	ans++;
	break;
      }
      i--;
      ans+=(1<<i);
      now=dp[POS(now,i)];
    }
    printf("%d\n",ans);
  }
}

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