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;

int a[100005];
int dp[100005][50];
int main()
{
	int n;
	scanf("%d",&n);
	for(int i = 1;i <= n;i++)
		scanf("%d",&a[i]);
	int L,Q;
	scanf("%d%d",&L,&Q);
	for(int i = 1;i <= n;i++)
	{
		int id = (int) (upper_bound(a+1,a+1+n,a[i]+L)-a-1);
		if(a[i] + L >= a[n])
			dp[i][0] = n;
		else
			dp[i][0] = id;
	}
	
	for(int i = 1;i <= 30; i++)
		for(int j = 1;j <= n; j++)
			dp[j][i] = dp [dp[j][i-1]][i-1];
	while(Q--)
	{
		int L,R;
		scanf("%d%d",&L,&R);
		if(R<L)
			swap(L,R);
		int ans = 1;
		for(int i=30;i>=0;i--)
		{
			if(dp[L][i] < R)
			{
				ans = ans + (1<<i);
				L = dp[L][i];
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}