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 <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 1000010;
int n;
int a[N];
int L, q, ans, lst;
int dp[N][21];
int pow[21];
int main() {
	memset(dp, 0x3f, sizeof(dp));
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	scanf("%d", &L);
	scanf("%d", &q);
	pow[0] = 1;
	for (int i = 1; i <= n; i++) {
		pow[i] = pow[i - 1] << 1;
		lst = max(i, lst);
		for (int j = lst + 1; j <= n; j++) {
			if (a[j] - a[i] <= L) {
				lst = j;
			}
		}
		dp[i][0] = lst;
	}
	for (int j = 1; j <= 20; j++) {
		
		for (int i = 1; i <= n; i++) {
			if (dp[i][j - 1] < 0x3f3f3f3f) dp[i][j] = dp[dp[i][j - 1]][j - 1];
		}
	}
	while (q--) {
		int x, y;
		scanf("%d%d", &x, &y);
		if (x > y) swap(x, y);
		ans = 0;
		for (int i = 20; i >= 0; i--) {
			if (dp[x][i] < y) {
				x = dp[x][i];
				ans += pow[i];
			}
		}
		printf("%d\n", ans + 1);
	}
	return 0;
}