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 <cstdio>
#include <algorithm>
#include <string>

using namespace std;

int n, l, q, h[100010], dt[20][100010];

int main()
{
    scanf("%d", &n);
    for(int i = 1 ; i <= n ; i++ ) scanf("%d", h+i);
    scanf("%d%d", &l, &q);
    for(int i = 1 ; i <= n ; i++ )
        dt[0][i] = (int)(upper_bound(h+1, h+n+1, h[i]+l) - h) - 1;
    for(int j = 1 ; j < 20 ; j++ ) for(int i = 1 ; i <= n ; i++ )
        dt[j][i] = dt[j-1][dt[j-1][i]];
    while( q-- )
    {
        int a, b, cnt = 0, d;
        scanf("%d%d", &a, &b);
        if( a > b ) swap(a, b);
        for(int i = 19 ; i >= 0 ; i-- )
        {
            if( dt[i][a] < b )
            {
                cnt += (1<<i);
                a = dt[i][a];
            }
        }
        printf("%d\n", cnt+1);
    }
    return 0;
}
