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>
using namespace std;
const int maxn = 1e5 + 10;
const int maxb = 20;
int x[maxn], r[maxn][maxb];
int n, L;
int main()
{
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) scanf("%d", &x[i]);
	scanf("%d", &L); 
	int j = n;
	r[n][0] = n + 1;
	for(int i = n - 1; i; --i)
	{
		while(x[j] - x[i] > L) --j;
		r[i][0] = j;
	}
	for(int j = 1; j < maxb; ++j) 
		for(int i = 1; i <= n; ++i)
			r[i][j] = r[i][j - 1] == n + 1 ? n + 1 : r[r[i][j - 1]][j - 1];
	int q;
	scanf("%d", &q);
	while(q--)
	{
		int u, v, ans = 0;
		scanf("%d%d", &u, &v);
		if(u > v) swap(u, v);
	 	for(int j = maxb - 1; ~j; --j)
	 		if(r[u][j] <= v)
	 		{
	 			u = r[u][j];
	 			ans += 1 << j;
	 		}
	 	if(u < v) ++ans;
	 	printf("%d\n", ans);
	}
	return 0;
}