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 <iostream>
using namespace std;
#define int long long
const int Maxn=100005;
const int inf=0x3f3f3f3f3f3f3f3f;
int n,a[Maxn],L,q;
int nxt[Maxn],nxt2[Maxn][20];
void init()
{
	for (int l=1,r=1;l<=n && r<=n && a[r]-a[l]<=L;++l)
	{
		while (r<=n && a[r]-a[l]<=L) r++;
		r--;
		nxt[l]=r;
	}
	nxt[n]=inf;
	for (int i=1;i<=n;++i) nxt2[i][0]=nxt[i];
	for (int i=1;i<=19;++i)
	{
		for (int j=1;j<=n;++j)
		{
			if (nxt2[j][i-1]==inf) nxt2[j][i]=inf;
			else nxt2[j][i]=nxt2[nxt2[j][i-1]][i-1];
		}
	}
}
signed main()
{
	cin>>n;
	for (int i=1;i<=n;++i) cin>>a[i];
	cin>>L>>q;
	init();
	while (q--)
	{
		int l,r,ans=0;
		cin>>l>>r;
		if (l>r) swap(l,r);
		for (int i=19;i>=0;i--)
		{
			if (nxt2[l][i]<=r)
			{
				ans+=1LL<<i;
				l=nxt2[l][i];
			}
		}
		if (l!=r) ans++;
		cout<<ans<<endl;
	}
	return 0;
}