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 maxn=1e5+10;
int n,a,b,p[maxn],l,Q;
int f[maxn][35];
int main()
{
    scanf("%d",&n);
    for (int i=1; i<=n; i++) scanf("%d",&p[i]);
    scanf("%d",&l);
    for (int i=1;i<=n;i++)
     {
        int idx=upper_bound(p+1,p+1+n,p[i]+l)-p-1;
        if(p[i]+l>=p[n]) f[i][0]=n;
        else f[i][0]=idx;
    }
    for (int j=1; j<=30; j++)
        for (int i=1;i<=n;i++)
            f[i][j]=f[f[i][j-1]][j-1];
    scanf("%d",&Q);
    while(Q--)
    {
        scanf("%d%d",&a,&b);
        if(a>b) swap(a,b);
        int cur=a,ans=0;
        for (int i=30; i>=0; i--)
        {
            if(f[cur][i]<b)
            {
                ans+=(1<<i);
                cur = f[cur][i];
            }
        }
        printf("%d\n",ans+1);
    }
    return 0;
}