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 ll N=1e5+5,LGN=20;
ll n,L,a[N],nxt[N][LGN];
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    cin>>n;
    for(ll i=0;i<n;i++)cin>>a[i];
    cin>>L;
    for(ll i=0;i<n;i++)nxt[i][0]=lower_bound(a+i,a+n,a[i]+L+1)-(a+1);
    for(ll j=1;j<LGN;j++){
        for(ll i=0;i<n;i++)nxt[i][j]=nxt[nxt[i][j-1]][j-1];
    }
    ll q;cin>>q;
    for(ll x,y,ans=0;q>0;q--,ans=0){
        cin>>x>>y;
        x--;y--;
        if(x>y)swap(x,y);
        for(ll i=LGN-1;i>=0;i--){
            if(nxt[x][i]<y)ans|=(1ll<<i),x=nxt[x][i];
        }
        cout<<ans+1<<"\n";
    }
}