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;
int N, L;
int x[101010];
int sparse[18][101010];
int main()
{
    scanf("%d", &N);
    for(int i=0; i<N; ++i) scanf("%d", x+i);
    x[N] = (int)2.1e9;
    scanf("%d", &L);
    
    for(int i=0; i<N; ++i)
        sparse[0][i] = upper_bound(x, x+N, x[i]+L) - 1 - x;
    for(int j=1; j<=17; ++j)
        for(int i=0; i<N; ++i)
            sparse[j][i] = sparse[j-1][sparse[j-1][i]];
    
    int Q; scanf("%d", &Q);
    for(int i=0; i<Q; ++i)
    {
        int a, b; scanf("%d%d", &a, &b); --a; --b;
        if(a>b) swap(a, b);
        int ans = 0;
        for(int j=17; j>=0; --j)
        {
            if(sparse[j][a] < b)
            {
                ans += (1<<j);
                a = sparse[j][a];
            }
        }
        printf("%d\n", ans+1);
    }
    return 0;
        
}