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 <bits/stdc++.h>
#define sc(x) scanf("%d", &x)
#define pr(x) printf("%d\n", x)
#define lowbit(x) (x & -x)
#define ll long long
#define N 100010
#define memset(x) memset(x, 0, sizeof(x))
using namespace std;
int n, a[N], f[N][35], l, q;
int main()
{
	//freopen("trav.in", "r", stdin);
	//freopen("trav.out", "w", stdout);
	sc(n);
	for (int i = 1; i <= n; i++)
	sc(a[i]);
	sc(l);
	for (int i = 1; i <= n; i++)
	{
		int id = (int)(upper_bound(a + 1, a + n + 1, a[i] + l) - a - 1);
		f[i][0] = id;
	}
	for (int j = 1; j <= 30; ++j)
	for (int i = 1; i <= n; ++i)
	f[i][j] = f[f[i][j - 1]][j - 1];
	sc(q);
	while (q--)
	{
		int x, y;
		sc(x);
		sc(y);
		if (x > y) swap(x, y);
		int sum = 1;
		for (int i = 30; i >= 0; i--)
		if (f[x][i] < y)
		{
			sum += 1 << i;
			x = f[x][i];
		}
		printf("%d\n", sum);
	}
	return 0;
}
