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 = 100005;

int x[N], p[N], d[N], pa[N][20];
int n, l, q, a, b;
vector<int> G[N];

void dfs(int v, int depth) {
	d[v] = depth;
	pa[v][0] = p[v];
	for(int k=1;k<20;k++)
		pa[v][k] = pa[pa[v][k-1]][k-1];
	for(int ch:G[v])
		dfs(ch, depth+1);
}

int main()
{
	cin >> n;
	for(int i=0;i<n;i++)
		cin >> x[i];
	cin >> l >> q;
	p[n-1]=n-1;
	for(int i=0;i<n-1;i++) {
		p[i] = upper_bound(x+i, x+n, x[i]+l)-x-1;
		G[p[i]].push_back(i);
	}
	dfs(n-1, 0);
	while(q--) {
		int ct = 0;
		cin >> a >> b;
		a--, b--;
		if(a > b) swap(a, b);
		for(int k=19;k>=0;--k){
			if(pa[a][k] < b)
				a = pa[a][k], ct += 1 << k;
		}
		cout << ++ct << endl;
	}
}
