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;
typedef long double ld;
typedef unsigned long long ull;

const int N = 100005;
const int LGN = 18;
int x[N], nxt[N][LGN];

int main(){
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	cout<<setprecision(32);

	int n; cin>>n;
	for(int i = 1; i <= n; i++){
		cin>>x[i];
	}
	int l; cin>>l;
	for(int i = 1; i <= n; i++){
		nxt[i][0] = upper_bound(x + 1, x + n + 1, x[i] + l) - x - 1;
	}
	for(int k = 1; k < LGN; k++){
		for(int i = 1; i <= n; i++){
			nxt[i][k] = nxt[nxt[i][k - 1]][k - 1];
		}
	}
	int q; cin>>q;
	while(q--){
		int a, b;
		cin>>a>>b;
		if(a > b)swap(a, b);
		int ans = 1;
		for(int i = LGN - 1; i >= 0; i--){
			if(nxt[a][i] < b){
				ans += (1<<i);
				a = nxt[a][i];
			}
		}
		cout<<ans<<'\n';
	}

	return 0;
}