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 <cstdio>
#include <algorithm>
#define ll long long
using namespace std;
inline int read(){
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') {x=(x<<1)+(x<<3)+(c^48);c=getchar();}
	return x*f;
}
const int maxn(1e5+5);
ll n,l,q;
ll s[maxn],f[maxn][35];

void swap(ll &a,ll &b){int t=a;a=b;b=t;}
void init(){
	n=read();
	for (int i(1);i<=n;++i) s[i]=read();
	l=read(),q=read();
	for (int i(1);i<=n;++i) f[i][0]=upper_bound(s+1,s+n+1,s[i]+l)-s-1;
	for (int i(1);i<=30;++i)
		for (int j(1);j<=n;++j)
			f[j][i]=f[f[j][i-1]][i-1];
}

signed main(){
	init();
	while (q--){
		int a=read(),b=read(),tot=0;
		if (a>b) swap(a,b);
		int v=a;
		for (int i=30;~i;--i){
			if (f[v][i]<b){
				tot+=1<<i;
				v=f[v][i];
			}
		}
		printf("%d\n",tot+1);
	}
	return 0; 
}
