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
#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int pat[20][100001];
int main()
{
	int num;
	scanf("%d", &num);
	vector<int>vec;
	for (int i = 0; i < num; i++)
	{
		int z;
		scanf("%d", &z);
		vec.push_back(z);
	}
	int gen;
	scanf("%d", &gen);
	for (int i = 0; i < num; i++)
	{
		int upp = upper_bound(vec.begin(), vec.end(), vec[i] + gen) - vec.begin() - 1;
		pat[0][i] = upp;
	}
	for (int i = 1; i < 20; i++)
	{
		for (int j = 0; j < num; j++)
		{
			pat[i][j] = pat[i - 1][pat[i - 1][j]];
		}
	}
	int query;
	scanf("%d", &query);
	for (int q = 0; q < query; q++)
	{
		int za, zb;
		scanf("%d%d", &za, &zb);
		za--, zb--;
		if (za > zb)swap(za, zb);
		int ans = 0;
		for (int i = 19; i >= 0; i--)
		{
			if (pat[i][za] < zb)
			{
				za = pat[i][za];
				ans += 1 << i;
			}
		}
		printf("%d\n", ans + 1);
	}
}