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;

typedef long long ll;
const int MAX_N = 100030;    // Change as necessary

int A[20][MAX_N];
int x[MAX_N];

int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  
  int n; cin >> n;
  for(int i=0;i<n;i++)
    cin >> x[i];
  int L,Q; cin >> L >> Q;
  
  x[n] = x[n-1]+1;
  n++;
  
  for(int i=0;i<n;i++)
    A[0][i] = upper_bound(x,x+n,x[i]+L)-x-1;
  
  for(int j=1;j<20;j++)
    for(int i=0;i<n;i++)
      A[j][i] = A[j-1][A[j-1][i]];

  while(Q--){
    int a,b; cin >> a >> b; a--; b--;
    if(a > b) swap(a,b);
    
    int ans = 0;
    for(int i=19;i>=0 && a < b;i--)
      if(A[i][a] <= b){
        a = A[i][a];
        ans += (1 << i);
      }
    if(a < b)
      ans++;
    cout << ans << "\n";
  }
  
  return 0;
}
