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>
#include<algorithm>
using namespace std;
int dp[21][200000];
int N,K,Q,x[200000],a,b;
int solve(int pos,int t){
	for(int i=19;i>=0;i--){
		if((t/(1<<i))%2==1)pos=dp[i][pos];
	}
	return pos;
}
int main(){
	cin>>N;for(int i=0;i<N;i++)cin>>x[i];cin>>K>>Q;
	for(int i=0;i<N;i++){
		int pos1=upper_bound(x,x+N,x[i]+K)-x;pos1--;
		dp[0][i]=pos1;
	}
	dp[0][N]=N;
	for(int i=0;i<20;i++){
		for(int j=0;j<N;j++)dp[i+1][j]=dp[i][dp[i][j]];
		dp[i+1][N]=N;
	}
	for(int i=0;i<Q;i++){
		cin>>a>>b;if(a>b)swap(a,b);a--;b--;
		int L=0,R=200000,M;
		while(true){
			M=(L+R)/2;
			int p1=solve(a,M-1),p2=solve(a,M);
			if(p1<b && p2>=b)break;
			if(p1>=b)R=M;
			if(p2<b)L=M;
		}
		cout<<M<<endl;
	}
	return 0;
}
