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;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rrep(i,n) for(int i = n-1;i >= 0;i--)
#define ALL(obj) begin(obj), end(obj)
int n, h, w,l,q,a,b;
int main(void) {
    cin >> n;vector<int> x(n); vector<vector<int>> dob(n+1, vector<int>(22));
    rep(i, n) cin >> x[i];
    cin >> l >> q;
    rep(i, n) dob[i][0] = upper_bound(ALL(x),x[i] + l) - x.begin() - 1;// 1日で行け最大地点
    rep(i, 20) rep(j, n) dob[j][i + 1] = dob[dob[j][i]][i]; // doubling
    rep(i, q) {
        cin >> a >> b; a--; b--;
        if (a > b) swap(a, b);// a> b のとき a<bでも問題ない
        int ans = 0;
        rrep(j, 21) if (dob[a][j] < b) a = dob[a][j], ans += (1 << j);
        ans++; cout << ans << endl;
    }
}