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>
#define rep(i,n) for(int i=0;i<n;++i)
using namespace std;
using ll=long long;
constexpr ll mod=1000000007;
constexpr int inf=1000000000;
constexpr ll linf=1000000000000000000;

int main(){
    int n;
    cin>>n;
    int x[n];
    rep(i,n)cin>>x[i];
    int l,q;
    cin>>l>>q;
    int r[20][n];
    rep(i,n){
        int to=upper_bound(x,x+n,x[i]+l)-x;
        r[0][i]=to-1;
    }
    for(int i=1;i<20;++i){
        for(int j=0;j<n;++j){
            r[i][j]=r[i-1][r[i-1][j]];
        }
    }
    rep(i,q){
        int a,b;
        cin>>a>>b;
        --a,--b;
        if(a>b)swap(a,b);
        int t=0;
        for(int j=19;j>=0;--j){
            if(r[j][a]<b){
                a=r[j][a];
                t+=(1<<j);
            }
        }
        cout<<t+1<<endl;
    }
}
