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;

const int MAXN = 1e5+10;
int dp[30][MAXN],x[MAXN],l,n;
int q;

int main()
{
    int i,j;
    cin >> n;
    for(int i = 1; i <= n; ++i)
        cin >> x[i];
    cin >> l;
    for(int i = 1; i < n; ++i)
        dp[0][i] = upper_bound(x+1,x+n+1,x[i]+l) - x - 1;
    for(i = 1; i < 30; ++i)
    {
        for(j = 1; j <= n; ++j)
            dp[i][j] = dp[i-1][dp[i-1][j]];
    }
    cin >> q;
    int a,b,res;
    while(q--)
    {
        res = 0;
        cin >> a >> b;
        if(a > b) swap(a,b);
        for(i = 29; i >= 0; --i)
        {
            if(dp[i][a] == 0) continue;
            if(dp[i][a] < b)
            {
                a = dp[i][a];
                res += (1<<i);
            }
        }
        cout << res+1 << endl;
    }
    return 0;
}
