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 <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <map>
#include <math.h>
#include <vector>
#include <queue>
using namespace std;
#define mod 1000000007
#define INF 0x7fffffffff
#define N 300100
#define endl '\n'
int a[110000],dp[110000][25];
int main()
{
    int n,m,L;
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    scanf("%d%d",&L,&m);
    for(int i=1;i<=n;i++) dp[i][0]=upper_bound(a+i+1,a+1+n,a[i]+L)-a-1;
    for(int j=0;j<18;j++)
        for(int i=1;i<=n;i++)
            dp[i][j+1]=dp[dp[i][j]][j];
    while(m--)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        if(l>r) swap(l,r);
        int ans=1;
        for(int i=17;i>=0;i--)
        {
            if(dp[l][i]<r) ans+=(1<<i),l=dp[l][i];
        }
        cout<<ans<<endl;
    }
	return 0;
}
