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;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	int n;
	cin >> n;
	vector<int> x(n);
	for(int i = 0; i < n; ++i) {
		cin >> x[i];
	}
	int L, q;
	cin >> L >> q;
	x.emplace_back(x.back() + L + 1);
	vector<vector<int> > p;
	p.resize(n);
	for(int i = 0; i < n; ++i) {
		p[i].resize(20);
	}
	for(int i = 0; i < n; ++i) {
		auto it = upper_bound(x.begin(), x.end(), x[i] + L);
		p[i][0] = (--it - x.begin()); 
	}
	for(int j = 1; j < 20; ++j) {
		for(int i = 0; i < n; ++i) {
			p[i][j] = p[p[i][j - 1]][j - 1];
		}
	}
	while(q--) {
		int a, b;
		cin >> a >> b;
		if(a > b) {
			swap(a, b);
		}
		--a;
		--b;
		int ans = 0;
		for(int j = 19; ~j; --j) {
			if(x[p[a][j]] < x[b]) {
				ans += 1 << j;
				a = p[a][j];
			}
		}
		cout << ans + 1 << '\n';
	}
	return 0;
}