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 N = 100020;
int n,q,f[N][30]; long long x[N],L;
int main(void)
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%lld",&x[i]);
    scanf("%lld\n%d",&L,&q);
    for (int i=1;i<=n;i++)
        f[i][0] = upper_bound( x+1 , x+n+1 , x[i] + L ) - x - 1;
    for (int k=1;k<=25;k++)
        for (int i=1;i<=n;i++)
            f[i][k] = f[f[i][k-1]][k-1];
    for (int i=1;i<=q;i++)
    {
        int a , b , ans = 0;
        scanf("%d%d",&a,&b);
        if ( a > b ) swap( a , b );
        for (int k=25;k>=0;k--)
            if ( f[a][k] < b ) a = f[a][k] , ans += 1<<k;
        if ( a < b ) a = f[a][0] , ans += 1;
        printf("%d\n",ans);
    }
    return 0;
}
