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 ll long long
#define rint register int
using namespace std;
const int N=1e5+50;
int n,a[N<<1],Q,f[35][N],L;
int main()
{
//freopen(".in","r",stdin);
//freopen(".out","w",stdout);
  scanf("%d",&n);
  for(int i=1;i<=n;i++) scanf("%d",&a[i]); 
  scanf("%d",&L);
  for(int i=1;i<=n;i++)
    {
        int x=upper_bound(a+1,a+n+1,a[i]+L)-a-1;
        f[0][i]=x;
    }
  for(int j=1;j<=30;j++)
    for(int i=1;i<=n;i++)
      f[j][i]=f[j-1][f[j-1][i]];
  scanf("%d",&Q);
  while(Q--)
  {
  	int l,r,ans=0;
  	scanf("%d%d",&l,&r);
  	if(l>r) swap(l,r); 
  	for(int i=30;i>=0;i--)
  	{  
	  if(f[i][l]<r) l=f[i][l],ans+=(1<<i);	
  	}
  	printf("%d\n",ans+1);
  }
  return 0;
}
