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;

#define SZ(v) ((int)(v).size())
#define ALL(v) (v).begin(),(v).end()
#define one first
#define two second
typedef long long ll;
typedef pair<int, int> pi;
const int INF = 0x3f2f1f0f;
const ll LINF = 1ll * INF * INF;

const int MAX_N = 1e5 + 10, LOG_N = 20;

int N, L, Q, Nr[MAX_N];
int P[LOG_N][MAX_N];
int main() {
	cin >> N; for(int i=1; i<=N; i++) scanf("%d", &Nr[i]); cin >> L >> Q;
	for(int l=1, r=1; l<=N; l++) {while(r+1<=N && Nr[r+1]-Nr[l] <= L) r++; P[0][l] = r;}
	for(int p=1; p<LOG_N; p++) for(int i=1; i<=N; i++) P[p][i] = P[p-1][P[p-1][i]];
	while(Q--) {
		int x, y; scanf("%d%d", &x, &y);
		if(x > y) swap(x, y);
		int ans = 0;
		for(int p=LOG_N-1; p>=0; p--) if(P[p][x] < y) ans += (1<<p), x = P[p][x];
		printf("%d\n", ans+1);
	}
	return 0;
}
