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;
signed main(){
  int n;
  cin>>n;
  int x[n];
  for(int i=0;i<n;i++) cin>>x[i];
  int l;
  cin>>l;
  vector<int> db[31];
  for(int i=0;i<31;i++) db[i].resize(n);
  for(int i=0,j=0;i<n;i++){
    while(j+1<n&&x[j+1]<=x[i]+l) j++;
    db[0][i]=j;
  }
  for(int j=0;j<31-1;j++)
    for(int i=0;i<n;i++)
      db[j+1][i]=db[j][db[j][i]];
  function<int(int,int)> calc=[&](int a,int b){
    if(b<=a) return 0;
    for(int j=0;j<40;j++){
      if(b<=db[j+1][a])
	return calc(db[j][a],b)+(1<<j);
    }
  };
  int q;
  cin>>q;
  for(int i=0;i<q;i++){
    int a,b;
    cin>>a>>b;
    a--;b--;
    if(a>b) swap(a,b);
    cout<<calc(a,b)<<endl;
  }
  return 0;
}
