Problem:
Score : 700 points

Problem Statement

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.

Partial Score

200 points will be awarded for passing the test set satisfying N \leq 10^3 and Q \leq 10^3.

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.

Sample Input 1

9
1 3 6 13 15 18 19 29 31
10
4
1 8
7 3
6 7
8 5

Sample Output 1

4
2
1
2

For the 1-st query, he can travel from the 1-st hotel to the 8-th hotel in 4 days, as follows:

Day 1: Travel from the 1-st hotel to the 2-nd hotel. The distance traveled is 2.

Day 2: Travel from the 2-nd hotel to the 4-th hotel. The distance traveled is 10.

Day 3: Travel from the 4-th hotel to the 7-th hotel. The distance traveled is 6.

Day 4: Travel from the 7-th hotel to the 8-th hotel. The distance traveled is 10.

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define ll long long
#define ull unsigned long long
#define F first
#define S second
#define eps 1e-6
#define mp make_pair
const long long INF=1e18+2;
const int MX=100004;
int N, M, k;
ll X[MX];
int dp[MX][18];
int get(int u, int D)
{
    if(D == 0) return u;
    for(int i=17;i>=0;i--)
    {
        if(D&(1<<i)) u=dp[u][i];
    }
    return u;
}
int main()
{
    memset(dp, -1, sizeof dp);
    scanf("%d", &N);
    for(int i=1;i<=N;i++) scanf("%d", &X[i]);
    scanf("%d", &k);
    for(int i=2;i<=N;i++)
    {
        dp[i][0]=lower_bound(X+1, X+1+N, X[i]-k)-X;
        for(int j=1;j<=17;j++)
        {
            dp[i][j]=dp[dp[i][j-1]][j-1];
        }
    }
    int q;
    scanf("%d", &q);
    while(q--)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        if(a > b) swap(a, b);
        int lo=0, hi=b-a, R=0, mid;
        while(lo <= hi)
        {
            mid=(lo+hi)>>1;
            int res=get(b, mid);
            if(res <= a)
            {
                R=mid;
                hi=mid-1;
            }
            else lo=mid+1;
        }
        cout<<R<<"\n";
    }
    return 0 ;
}