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;
ll n, l, q, a, b, v[100010][32];
vector<ll> x;

int main(){
//	cin.tie(0);
//	ios::sync_with_stdio(false);
	cin >> n;
	x.resize(n+2);
	for(int i=1; i<=n; ++i) cin >> x[i];
	x[0] = -1;
	x[n+1] = 1e18+9;
	cin >> l;
	for(int i=1; i<=n; ++i)
		v[i][0] = upper_bound(x.begin(), x.end(), x[i]+l) - x.begin() - 1;
	for(int i=1; i<20; ++i)
		for(int j=1; j<=n; ++j)
			v[j][i] = v[v[j][i-1]][i-1];
//	for(int i=1; i<=n; ++i)
//		cout << i << " " << v[i][0] << "\n";
//	for(int i=1; i<2; ++i)
//		for(int j=0; j<20; ++j)
//			cout << v[i][j] << (j == 19 ? "\n" : " ");
	cin >> q;
	while(q--){
		cin >> a >> b;
		ll ans = 0;
		if(a > b) swap(a, b);
		for(int i=19; i>=0; --i){
			if(v[a][i] < b){
				a = v[a][i];
				ans += 1 << i;
			}
		}
		cout << ans + 1 << "\n";
	}
}