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;

const int K = 20;
int n, q, l;
int x[100000];
int d[K][100000];
void pre() {
  int j = 0;
  for (int i = 0; i < n; ++i) {
    while (j < n && x[j] - x[i] <= l) {
      ++j;
    }
    d[0][i] = j - 1;
  }
  
  for (int k = 1; k < K; ++k) {
    for (int i = 0; i < n; ++i) {
      d[k][i] = d[k - 1][d[k - 1][i]];
    }
  }
}

int sub(int a, int b) {
  int cnt = 0;
  int i = a;
  for (int k = K - 1; k >= 0; --k) {
    int ni = d[k][i];
    if (ni < b) {
      cnt += 1 << k;
      i = ni;
    }
  }
  return cnt + 1;
}

int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> x[i];
  cin >> l >> q;
  pre();
  
  for (int j = 0; j < q; ++j) {
    int a, b;
    cin >> a >> b;
    --a; --b;
    if (a > b) swap(a, b);
    cout << sub(a, b) << endl;
  }
}