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 
using namespace std;
const int maxn=110000;
ll a[maxn];
int n;
ll l;
ll q;
int f[maxn][35]={};
inline void init()
{
  scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",a+i); scanf("%d%d",&l,&q);
  for(int i=1;i<=n;i++)
  {
    int x=(int)(upper_bound(a+1,a+n+1,a[i]+l)-a-1);
    f[i][0]=x;
  }
  for(int j=1;j<=30;j++) for(int i=1;i<=n;i++)  f[i][j]=f[f[i][j-1]][j-1];
}
int main()
{
	init();
	while(q--)
    {
      ll a,b;
      scanf("%lld %lld",&a,&b);
      if(a>b)
        swap(a,b);
      int ans=0;
      int cur=a;
      for(int i=30;i>=0;i--)
	    if(f[cur][i]<b)
	    {
	      ans+=1<<i;
	      cur=f[cur][i];
	    }
      printf("%d\n",ans+1);
    }
	return 0;
}
