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: Wrong Answer)

// Standard I/O
#include <iostream>
#include <sstream>
#include <cstdio>
// Standard Library
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cmath>
// Template Class
#include <complex>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
// Container Control
#include <algorithm>

using namespace std;

#define rep( i, n ) for( int i = 0; i < n; ++i )
#define irep( i, n ) for( int i = n-1; i >= 0; --i )
#define reep( i, s, n ) for ( int i = s; i < n; ++i )
#define ireep( i, n, s ) for ( int i = n-1; i >= s; --i )
#define foreach(itr, x) for( typeof(x.begin()) itr = x.begin(); itr != x.end(); ++itr)

#define mp make_pair
#define mt make_tuple
#define pb push_back
#define eb emplace_back
#define all( v ) v.begin(), v.end()
#define fs first
#define sc second
#define vc vector

// for visualizer.html
double SCALE = 1.0;
double OFFSET_X = 0.0;
double OFFSET_Y = 0.0;
#define LINE(x,y,a,b) cerr << "line(" << SCALE*(x) + OFFSET_X << ","	\
	<< SCALE*(y) + OFFSET_Y << ","										\
	<< SCALE*(a) + OFFSET_X << ","										\
	<< SCALE*(b) + OFFSET_Y << ")" << endl;
#define CIRCLE(x,y,r) cerr << "circle(" << SCALE*(x) + OFFSET_X << ","	\
	<< SCALE*(y) + OFFSET_Y << ","										\
	<< SCALE*(r) << ")" << endl;

typedef long long ll;
typedef complex<double> Point;

typedef pair<int, int> pii;
typedef pair<int, pii> ipii;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector< vector<int> > vii;
typedef vector< vector<double> > vdd;

typedef vector<int>::iterator vi_itr;

const int IINF = 1 << 28;
const int MOD = 1e9+7;
const double INF = 1e30;
const double EPS = 1e-10;
const double PI = acos(-1.0);

// Direction : L U R D
const int dx[] = { -1, 0, 1, 0};
const int dy[] = { 0, -1, 0, 1 };

int main()
{
	int N;
	ll x[100000], L;

	cin >> N;
	rep(i, N) cin >> x[i];
	cin >> L;

	ll cur = x[0];
	rep(i, N){
		ll org = x[i];
		x[i] = x[i] - cur;
		cur += x[i];
	}
	
	ll dp[100000] = { 0 };
	ll dist = 0;
	dp[0] = 1;
	reep(i, 1, N){
		if( dist + x[i] <= L ){
			dp[i] = dp[i-1];
			dist += x[i];
		}
		else{
			dp[i] = dp[i-1] + 1;
			dist = x[i];
		}
	}
	dp[0] = 0;
	
	int Q;
	cin >> Q;
	rep(i, Q){
		int a, b;
		cin >> a >> b;

		--a; --b;
		if( a > b ) swap(a, b);
		cout << max(dp[b] - dp[a], 1ll) << endl;
	}
}