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;
typedef long long ll;
const int maxn=2e5+10;
ll a[maxn];
ll f[maxn][35];
int main(){
    int n,step,q,i,j,p;
    scanf("%d",&n);
    for (i=0;i<n;i++)  scanf("%lld",&a[i]);
    scanf("%d%d",&step,&q);
    for (i=0;i<n;i++)  {
        p=upper_bound(a,a+n,a[i]+step)-a;
        f[i+1][0]=p;
    }
    for (int j=1;j<=30;j++)
        for (int i=1;i<=n;i++)
            f[i][j]=f[f[i][j-1]][j-1];
    int l,r;
    while (q--){
        scanf("%d%d",&l,&r);
        if (l>r) swap(l,r);
        ll ans=0;
        for (int i=30;i>=0;i--){
            if (f[l][i]<r){
                ans+=(ll)(1<<i);
                l=f[l][i];
            }
        }
        printf("%lld\n",ans+1);
    }
    return 0;
}