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: Runtime Error)

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

#define mod 1e9+7
#define ll long long int
#define epsilon 1e-7
int min(ll x, ll y){
    if(x < y)return x;
    else return y;
}
int compare_int(const void *c, const void *d){
    ll *C = (ll *)c;
    ll *D = (ll *)d;
    if(*C < *D)return 1;
    if(*C > *D)return -1;
    return 0;
}
int days(int start, int goal, int l, int n, int x[], int xx[n+1][n+1]){
    if(xx[start][goal]!=0)return xx[start][goal];
    int j = start;
    while(x[j] - x[start] <= l){
        if(j == goal)return 1;
        j++;
    }
    j--;
    xx[start][j] = 1;
    return xx[start][goal] = 1 + days(j, goal, l, n, x, xx);
}

int main(void){
    int n, l, q;
    scanf("%d", &n);
    int x[n+1];
    for(int i = 0;i < n;i++){
        scanf("%d", &x[i+1]);
    }
    scanf("%d %d", &l, &q);
    int a[q], b[q];
    for(int i = 0;i < q;i++){
        scanf("%d %d", &a[i], &b[i]);
    }
    int xx[n+1][n+1];
    for(int i = 0;i <= n;i++){
        for(int j = 0;j <= n;j++){
            xx[i][j] = 0;
        }
    }
    for(int i = 0;i < q;i++){
        if(a[i] > b[i]){
            int t = a[i];
            a[i] = b[i];
            b[i] = t;
        }
        printf("%d\n", days(a[i], b[i], l, n, x, xx));
    }
    return 0;
}