Problem: 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.

Examples

Input

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


Output

4
2


Input

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


Output

4
6
11
2
3

Solution (in CPP):

Solution:
// warm heart, wagging tail,and a smile just for you!
//                                                                     ███████████
//                                                                   ███╬╬╬╬╬╬╬╬╬╬███
//                                                                ███╬╬╬╬╬████╬╬╬╬╬╬███
//                                            ███████████       ██╬╬╬╬╬████╬╬████╬╬╬╬╬██
//                                      █████████╬╬╬╬╬████████████╬╬╬╬╬██╬╬╬╬╬╬███╬╬╬╬╬██
//                               ████████╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬█████████╬╬╬╬╬╬██╬╬╬╬╬╬╬██
//                             ████╬██╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬█████████╬╬╬╬╬╬╬╬╬╬╬██
//                           ███╬╬╬█╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬██╬╬███╬╬╬╬╬╬╬█████
//                         ███╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬██╬╬╬████████╬╬╬╬╬██
//                       ███╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬███╬╬╬╬╬╬╬╬╬███
//                     ███╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬█████╬╬╬╬╬╬╬██
//                 ████╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬████╬╬╬╬╬████
//     █████████████╬╬╬╬╬╬╬╬██╬╬╬╬╬████╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬█████╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬███╬╬╬╬██████
//   ████╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬██╬╬██████╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬██████╬╬╬╬╬╬╬███████████╬╬╬╬╬╬╬╬██╬╬╬██╬╬╬██
// ███╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬██╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬██╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬████╬╬╬╬╬╬╬╬╬╬╬█╬╬╬╬╬╬╬██╬╬╬╬╬╬╬╬██
// ██╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬██╬╬╬╬▓▓▓▓▓▓╬╬╬████╬╬████╬╬╬╬╬╬╬▓▓▓▓▓▓▓▓██╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬██╬╬╬╬╬╬╬███
// ██╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬██████▓▓▓▓▓▓▓╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬▓▓▓▓▓▓▓██╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬██╬╬╬╬█████
// ███╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬███╬╬╬╬╬██╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬█████╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬████████
//   ███╬╬╬╬╬╬╬╬╬╬╬╬╬█████╬╬╬╬╬╬╬╬██╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬███╬╬██╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬██
//       ██████████████  ████╬╬╬╬╬╬███████████████████████████╬╬╬╬╬██╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬████
//                         ███████                           █████  ███████████████████  
//
#include "bits/stdc++.h"
using namespace std;
#define INF (1<<30)
#define LINF (1LL<<60)
#define fs first
#define sc second
#define int long long
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define FOR2(i,a,b) for(int i=(a);i<=(b);++i)
#define RFOR(i,a,b) for(int i = (b-1);i>=(a);--i)
#define RFOR2(i,a,b) for(int i = (b);i>=(a);--i)
#define REP(i,n)  FOR(i,0,(n))
#define REP2(i,n)  FOR2(i,0,(n))
#define RREP(i,n) RFOR(i,0,(n))
#define RREP2(i,n) RFOR2(i,0,(n))
#define ITR(itr,mp) for(auto itr = (mp).begin(); itr != (mp).end(); ++itr)
#define RITR(itr,mp) for(auto itr = (mp).rbegin(); itr != (mp).rend(); ++itr)
#define range(i,a,b) ((a)<=(i) && (i)<(b))
#define range2(i,a,b) ((a)<=(i) && (i)<=(b))
#define debug(x)  cout << #x << " = " << (x) << endl
#define SP << " " << 
template<typename T1,typename T2> inline bool chmin(T1 &a,T2 b){if(a>b) {a=b; return true;} else return false;}
template<typename T1,typename T2> inline bool chmax(T1 &a,T2 b){if(a<b) {a=b; return true;} else return false;}
#define MSB(x) (63-__builtin_clzll(x))
#define pcnt(x) (__builtin_popcountll(x))
#define parity(i,j) (i&(1LL<<j))
typedef pair<int,int> P;
typedef tuple<int,int,int> T;
typedef vector<int> vec;
typedef vector<vector<int>> mat;

void solve(){
  int N,K;
  cin >> N >> K;

  vector<int> a(N);
  REP(i,N) cin >> a[i];

  mat L(30,vec(N)), R = L, Lsum = L, Rsum = L;
  REP(i,N){
    L[0][i] = lower_bound(a.begin(),a.end(),a[i]+K) - a.begin();
    R[0][i] = upper_bound(a.begin(),a.end(),a[i]-K) - a.begin() - 1;
    Lsum[0][i] = Rsum[0][i] = i;
  }

  REP(i,29){
    REP(j,N){
      if(L[i][j]==N) L[i+1][j] = N, Lsum[i+1][j] = Lsum[i][j];
      else L[i+1][j] = L[i][L[i][j]], Lsum[i+1][j] = Lsum[i][j]+Lsum[i][L[i][j]];

      if(R[i][j]==-1) R[i+1][j] = -1, Rsum[i+1][j] = Rsum[i][j];
      else R[i+1][j] = R[i][R[i][j]], Rsum[i+1][j] = Rsum[i][j]+Rsum[i][R[i][j]];
    }
  }
  
  int Q; cin >> Q;
  REP(_,Q){
    int l,r;
    cin >> l >> r; l--; r--;
    int now = l, mx = 1, sum = 0;
    RREP(i,30){
      if(L[i][now] <= r) mx += (1<<i), sum += Lsum[i][now], now = L[i][now];
    }
    sum += now;
    now = r; 
    RREP(i,30){
      if(R[i][now] >= l) sum -= Rsum[i][now], now = R[i][now];
    }
    sum -= now;
    cout << abs(sum) + mx << endl;
  }
}

signed main(){
  ios::sync_with_stdio(false);
  cin.tie(0);

  int T = 1;
  // cin >> T;

  while(T--) solve();

  return 0;
}