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<iostream>
#include<cstdio>
using namespace std;
const int magic=307;
int n,x[111111],l,q,a,b,nxt[111111],bignxt[111111],pos,ans;
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&x[i]);
	scanf("%d",&l);
	pos=n;
	for (int i=n;i>=1;i--)
	{
		while(pos>=1 && x[pos]-x[i]>l) pos--;
		nxt[i]=pos;
	}
	for (int i=1;i<=n;i++)
	{
		pos=i;
		for (int j=1;j<=magic;j++)
		{
			pos=nxt[pos];
		}
		bignxt[i]=pos;
	}
	scanf("%d",&q);
	while(q--)
	{
		scanf("%d%d",&a,&b);
		if (a>b) swap(a,b);
		ans=0;
		while(bignxt[a]<b) 
		{
			ans+=magic;
			a=bignxt[a];
		}
		while(nxt[a]<b) 
		{
			ans++;
			a=nxt[a];
		}
		if (a!=b) ans++;
		printf("%d\n",ans);
	}
	return 0;
}