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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
const int maxn = int(1e5) + 7, inf = 0x3f3f3f3f;
int n, x[maxn], up[maxn][20], l, q, u, v, e[27] = {1}, ans, tmp;
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", x + i);
    scanf("%d%d", &l, &q);
//    memset(up, 0x3f, sizeof(up));
    int cur = 1;
    for (int i = 1; i <= n; i++) while (x[i] > x[cur] + l) up[cur++][0] = i - 1;//, printf("asd %d\n",up[cur-1][0]);
    for (int j = 1; j < 20; j++) {
        e[j] = e[j - 1] << 1;
        for (int i = 1; i <= n; i++) if (up[i][j - 1] ) //if (up[i][j - 1] < inf)
            up[i][j] = up[up[i][j - 1]][j - 1];
    }
    while (ans = 0, q--) {
        scanf("%d%d", &u, &v);
        if (u > v) std::swap(u, v);
        for (int i = 19; i >= 0; i--) if (up[u][i] < v && up[u][i]) u = up[u][i], ans += e[i];
        printf("%d\n", ans + 1);
    }
    return 0;
}//倍增优化DP？