Problem:
Score : 800 points

Problem Statement

There are N points on a number line, i-th of which is placed on coordinate X_i.
These points are numbered in the increasing order of coordinates.
In other words, for all i (1 \leq i \leq N-1), X_i < X_{i+1} holds.
In addition to that, an integer K is given.

Process Q queries.

In the i-th query, two integers L_i and R_i are given.
Here, a set s of points is said to be a good set if it satisfies all of the following conditions.
Note that the definition of good sets varies over queries.

Each point in s is one of X_{L_i},X_{L_i+1},\ldots,X_{R_i}.

For any two distinct points in s, the distance between them is greater than or equal to K.

The size of s is maximum among all sets that satisfy the aforementioned conditions.

For each query, find the size of the union of all good sets.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq 10^9

0 \leq X_1 < X_2 < \cdots < X_N \leq 10^9

1 \leq Q \leq 2 \times 10^5

1 \leq L_i \leq R_i \leq N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
X_1 X_2 \cdots X_N
Q
L_1 R_1
L_2 R_2
\vdots
L_Q R_Q

Output

For each query, print the size of the union of all good sets in a line.

Sample Input 1

5 3
1 2 4 7 8
2
1 5
1 2

Sample Output 1

4
2

In the first query, you can have at most 3 points in a good set.
There exist two good sets: \{1,4,7\} and \{1,4,8\}.
Therefore, the size of the union of all good sets is |\{1,4,7,8\}|=4.

In the second query, you can have at most 1 point in a good set.
There exist two good sets: \{1\} and \{2\}.
Therefore, the size of the union of all good sets is |\{1,2\}|=2.

Sample Input 2

15 220492538
4452279 12864090 23146757 31318558 133073771 141315707 263239555 350278176 401243954 418305779 450172439 560311491 625900495 626194585 891960194
5
6 14
1 8
1 13
7 12
4 12

Sample Output 2

4
6
11
2
3

Solution in C++ (GCC 9.2.1 with AC Library v1.1) (judge verdict: Accepted)

////////////////////////////////////////////////////////////////////
////////////        AUTHOR : LYNMISAKURA        ////////////////////
////////////////////////////////////////////////////////////////////
#include <bits/stdc++.h>
using namespace std;
 
#define REP(i,n) for(int i = 0;i < n;i++)
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(),(x).end()

template<class T> bool chmax(T& a,T b){ if(a < b){ a = b; return true; }else return false; }
template<class T> bool chmin(T& a,T b){ if(a > b){ a = b; return true; }else return false; }
 
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vl = vector<ll>;
using Pi = pair<int,int>;
using Pl = pair<ll,ll>;
using vpi = vector<Pi>;
using vpl = vector<Pl>;
 
#define debug(arr) cout << #arr << " = " << arr << '\n'
#define debug2(a,b) cout << "[" << #a << "," << #b << "] = " << "[" << a << "," << b << "]" << '\n'
template<class T> ostream &operator << (ostream& out, const vector<T>& arr) {
	cout << "{"; for (int i = 0; i < arr.size(); i++)cout << (!i ? "" : ", ") << arr[i]; cout << "}";
	return out;
}
template<class T> ostream &operator << (ostream& out, const vector<vector<T> >& arr) {
	cout << "{\n"; for (auto& vec : arr)cout << "  " << vec << ",\n"; cout << "}";
	return out;
}
template<class S,class T> ostream &operator << (ostream& out, const pair<S,T>& p){
  cout << "{" << p.first << "," << p.second << "}" << '\n';
  return out;
}
template<class T> istream &operator >> (istream& in, vector<T>& arr) {
	for (auto& i : arr)cin >> i; return in;
}


#define maxn 200020
#define maxlog 30
const ll INF = (1LL << 55);
int N,K,Q;
ll X[maxn];
int R[maxn][maxlog],L[maxn][maxlog];
ll RSum[maxn][maxlog],LSum[maxn][maxlog];
int get_l(int x,int k){
  int tmp = x;
  for(int i = 0;i < maxlog;i++){
    if(k >> i & 1){
      tmp = L[tmp][i];
    }
  }
  return tmp;
}
int get_r(int x,int k){
  int tmp = x;
  for(int i = 0;i < maxlog;i++){
    if(k >> i & 1){
      tmp = R[tmp][i];
    }
  }
  return tmp;
}
ll get_ls(int x,int k){
  int tmp = x;
  ll res = x;
  for(int i = 0;i < maxlog;i++){
    if(k >> i & 1){
      res += LSum[tmp][i];
      tmp = L[tmp][i];
    }
  }
  return res;
}
ll get_rs(int x,int k){
  int tmp = x;
  ll res = x;
  for(int i = 0;i < maxlog;i++){
    if(k >> i & 1){
      res += RSum[tmp][i];
      tmp = R[tmp][i];
    }
  }
  return res;
}
void init(){
  R[0][0] = 1;
  R[N+1][0] = N+1;
  L[N+1][0] = N;
  L[0][0] = 0;
  int l = 1,r = 2;
  for(;l <= N;l++){
    for(;r <= N+1;r++){
      if(X[l] + K <= X[r]){
        R[l][0] = r;
        break;
      }
    }
  }
  for(int i = 1;i < maxlog;i++){
    for(int j = 0;j <= N+1;j++){
      R[j][i] = R[R[j][i-1]][i-1];
    }
  }
  l = N-1,r = N;
  for(;r >= 1;r--){
    for(;l >= 0;l--){
      if(X[l] + K <= X[r]){
        L[r][0] = l;
        break;
      }
    }
  }
  for(int i = 1;i < maxlog;i++){
    for(int j = 0;j <= N+1;j++){
      L[j][i] = L[L[j][i-1]][i-1];
    }
  }
  for(int i = 1;i <= N;i++){
    RSum[i][0] = R[i][0];
    LSum[i][0] = L[i][0];
  }
  for(int i = 1;i < maxlog;i++){
    for(int j = 1;j <= N;j++){
      RSum[j][i] = RSum[j][i-1] + RSum[R[j][i-1]][i-1];
      LSum[j][i] = LSum[j][i-1] + LSum[L[j][i-1]][i-1];
    }
  }
}

ll query(int l,int r){
  
  int low = 0,high = N;
  while(low + 1 < high){
    int mid = (low + high)/2;
    if(get_r(l,mid) <= r){
      low = mid;
    }else high = mid;
  }
  
  int step = low;
  //cout << get_ls(r,step) << ' ' << get_rs(l,step) << endl;
  return get_ls(r,step) - get_rs(l,step) + step + 1;
  
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  cin >> N >> K;
  for(int i = 0;i <= N+1;i++){
    if(i == 0) X[i] = -INF;
    else if(i <= N) cin >> X[i];
    else X[i] = INF;
  }
  
  init();
   
  cin >> Q;
  for(int i = 0;i < Q;i++){
    int l,r;cin >> l >> r;
    cout << query(l,r) << '\n';
  }
  
}