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;
typedef long long LL;
const int maxn = 1e5 + 5;
int dp[maxn][20];
int x[maxn];

int main(){
    int N, L, Q, a, b;
    scanf("%d", &N);
    for(int i = 1; i <= N; i++){
        scanf("%d", &x[i]);
    }
    scanf("%d", &L);
    x[0] = x[1];
    for(int i = 1, l = 0; i <= N; i++){
        while(x[i] - x[l] > L) l++;
        dp[i][0] = l;
        for(int j = 1; j < 20; j++){
            dp[i][j] = dp[dp[i][j - 1]][j - 1];
        }
    }
    scanf("%d", &Q);
    while(Q--){
        scanf("%d%d", &a, &b);
        if(a < b) swap(a, b);
        int ans = 0;
        for(int d = 19; d >= 0; d--){
            if(dp[a][d] > b){
                ans += (1 << d);
                a = dp[a][d];
            }
        }
        printf("%d\n", ans + 1);
    }
    return 0;
}
