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 N = (int)1e5 + 100;
const int lg = 17;
int x[N],d[N][lg + 1];
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n,L;
	cin >> n;
	for(int i = 1;i <= n;i++)cin >> x[i];
	cin >> L;
	for(int i = n;i >= 1;--i){
	    d[i][0] = ((upper_bound(x + 1,x + 1 + n, x[i] + L)) - x - 1);
	    for(int j = 1;j <= lg;++j){
	        d[i][j] = d[d[i][j - 1]][j - 1];
	    }
	}
	int q;
	cin >> q;
	while(q--){
	    int a,b;
	    cin >> a >> b;
	    if(a == b){
	        cout << 0 << '\n';
	        continue;
	    }
	    if(a > b)swap(a, b);
	    int ans = 1;
	    for(int i = lg;i >= 0;--i){
	        if(d[a][i] < b){
	            a = d[a][i];
	            ans += (1 << i);
	        }
	    }
	    cout << ans << '\n';
	}
	
  	return 0;
} 