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<iostream>
using namespace std;
typedef long long ll;
int dp[18][100000];

int main(){
	ll n; cin >> n;
	ll x[n];
	ll sum = 0;
	for(int i = 0; i < n; ++i){
		cin >> x[i];
		for(int j = 0; j < 18; ++j){
			dp[j][i] = 0;
		}
	}

	ll l, q; cin >> l >> q;
	/*for(int i = 0; i < n; ++i){
		if(x[i] > l){
			dp[0][0] = i - 1;
			break;
		}
	}
	for(int i = 1;;++i){
		ll day = (1 << i);
		ll lower = 0;
		ll upper = n - 1;
		ll oldmid = 0;
		ll oldupper = 0;
		ll oldlower = 0;
		ll dist = 0;
		while(lower != upper){
			ll mid = (upper + lower) / 2;
			ll need = 0;
			ll now = 0;
			while(1){
				if(now >= mid) break;
				if(dist + x[now + 1] - x[now] <= l){
					dist += x[now + 1] - x[now];
					++now;
				}else{
					++need;
					if(need == day) break;
					dist = 0;
				}
				//cout << mid << endl;
			}
			oldupper = upper;
			oldlower = lower;
			if(now >= mid) lower = mid;
			else upper = mid;
			if(oldupper == upper && oldlower == lower) break;
		}
		dp[i][0] = upper;
		if(upper >= n - 1){
			for(int j = i + 1; j < 18; ++j) dp[j][0] = upper;
			break;
		}
	}*/
	for(int i = 0; i < n; ++i){
		ll upper = n - 1;
		ll lower = 0;
		while(upper - lower > 1){
			ll mid = (upper + lower) / 2;
			if(x[mid] - x[i] <= l) lower = mid;
			else upper = mid;
		}
		dp[0][i] = (x[upper] - x[i] <= l ? upper : lower);
	}

	for(int i = 1; i < 18; ++i){
		for(int j = 0; j < n; ++j) dp[i][j] = dp[i - 1][dp[i - 1][j]];
	}

	//for(int i = 0; i < 18; ++i) cout << dp[i][0] << ", ";
	//cout << endl;
	for(int que = 0; que < q; ++que){
		ll a, b; cin >> a >> b;
		--a; --b;
		if(a > b) a ^= b ^= a ^= b;
		ll lower = 0;
		ll upper = 17;
		ll ans = 0;
		while(1){
			while(upper - lower > 1){
				ll mid = (upper + lower) / 2;
				if(dp[mid][a] >= b) upper = mid;
				else lower = mid;
			}
			if(dp[upper][a] == b){
				ans += (1 << upper);
				break;
			}else if(dp[lower][a] == b){
				ans += (1 << lower);
				break;
			}else{
				ll use = (dp[a][upper] < b ? upper : lower);
				ans += (1 << use);
				a = dp[use][a];
				upper = 17;
				lower = 0;
			}
		}
		cout << ans << endl;
	}
	return 0;
}