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;
#ifdef LOCAL_DEBUG
  #include "LOCAL_DEBUG.hpp"
#endif
#define int long long

signed main(){

  int n; cin >> n;
  vector<int> x(n);
  for(int i = 0; i < n; i++){
    cin >> x[i];
  }
  int l, q; cin >> l >> q;

  static int dp[100001][40];
  for(int i = 0; i < n; i++){
    int pos = upper_bound(x.begin(),x.end(),x[i] + l) - x.begin() - 1;
    dp[i][0] = pos;
  }
  for(int k = 0; k < 30; k++){
    for(int i = 0; i < n; i++){
      dp[i][k+1] = dp[dp[i][k]][k];
    }
  }

  for(int i = 0; i < q; i++){
    int a, b; cin >> a >> b;
    a--, b--;
    if(a > b) swap(a, b);
    int ans = 0;
    for(int k = 30; k >= 0; k--){
      if(dp[a][k] < b){
        a = dp[a][k];
        ans += 1 << k;
        if(b <= a) break;
      }
    }
    cout << ans + 1 << endl;
  }
  return 0;
}
