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.

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.

Example

Input

9
1 3 6 13 15 18 19 29 31
10
4
1 8
7 3
6 7
8 5


Output

4
2
1
2
import java.util.Arrays;
import java.util.Scanner;

public class Main {

	static Scanner sc = new Scanner(System.in);
	static int N, L, logN;
	static int[] X;
	static int[][] table;

	public static void main(String[] args) {
		N = sc.nextInt();
		X = new int[N];
		for (int i = 0; i < N; ++i) {
			X[i] = Integer.parseInt(sc.next());
		}
		L = sc.nextInt();
		logN = 1;
		while ((1 << logN) < N) {
			logN++;
		}
		build();
		int Q = sc.nextInt();
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < Q; ++i) {
			int A = Integer.parseInt(sc.next()) - 1;
			int B = Integer.parseInt(sc.next()) - 1;
			sb.append(solve(Math.min(A, B), Math.max(A, B)) + "\n");
		}
		System.out.print(sb);
	}

	static void build() {
		table = new int[logN + 1][N];
		for (int i = 0; i < N; ++i) {
			int pos = Arrays.binarySearch(X, X[i] + L);
			if (pos < 0) {
				pos = -pos - 2;
			}
			table[0][i] = pos;
		}
		for (int i = 1; i <= logN; ++i) {
			for (int j = 0; j < N; ++j) {
				int next = table[i - 1][j];
				table[i][j] = table[i - 1][next];
			}
		}
	}

	static int solve(int l, int r) {
		int len = 0;
		for (int i = 0; i <= logN; ++i) {
			if (table[i][l] < r) {
				++len;
			} else {
				break;
			}
		}
		if (len == 0) return 1;
		return (1 << (len - 1)) + solve(table[len - 1][l], r);
	}

}
