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() {
	long long N, L, Q;
	cin >> N;
	vector<long long> X(N);
	for (int i = 0; i < N; i++) cin >> X[i];
	cin >> L >> Q;
	long long K = log2(N);
	vector<vector<long long> > D(N, vector<long long>(K + 1));
	for (int i = 0; i < N; i++) {
		D[i][0] = upper_bound(X.begin(), X.end(), X[i] + L) - X.begin() - 1;
	}
	for (int j = 1; j <= K; j++) {
		for (int i = 0; i < N; i++) {
			D[i][j] = D[D[i][j - 1]][j - 1];
		}
	}
	for (int i = 0; i < Q; i++) {
		long long S, T;
		cin >> S >> T;
		S--, T--;
		if (S > T) swap(S, T);
		long long ANS = 0;
		for (int j = K; j >= 0; j--) {
			if (D[S][j] < T) {
				S = D[S][j];
				ANS += 1LL << j;
			}
		}
		cout << ANS + 1 << endl;
	}
}