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:
#include <iostream>
#include <string>
#include <vector>
#include <array>
#include <queue>
#include <deque>
#include <algorithm>
#include <set>
#include <map>
#include <bitset>
#include <cmath>
#include <functional>
#include <cassert>
#include <iomanip>
#define vll vector<ll>
#define vvvl vector<vvl>
#define vvl vector<vector<ll>>
#define VV(a, b, c, d) vector<vector<d>>(a, vector<d>(b, c))
#define VVV(a, b, c, d) vector<vvl>(a, vvl(b, vll (c, d)));
#define re(c, b) for(ll c=0;c<b;c++)
#define all(obj) (obj).begin(), (obj).end()
typedef long long int ll;
typedef long double ld;
using namespace std;


typedef array<ll, 2> ary;
pair<vvl, vvl> calc(ll n, ll k, vector<ll>& a){
  vvl nxt = VV(n, 30, -1, ll);
  vvl s = VV(n, 30, 0, ll);
  for(int i=0;i<30;i++){
    if(i==0){
      for(int j=0;j<n;j++){
        auto idx = lower_bound(all(a), a[j]+k) - a.begin();//k先のidx
        if(j==n-1) idx = n;
        nxt[j][0] = idx;
        if(idx!=n) s[j][0] = idx;
      }
    }else{
      for(int j=0;j<n;j++){
        nxt[j][i] = (nxt[j][i-1]==n?n:nxt[nxt[j][i-1]][i-1]);
        s[j][i] = s[j][i-1] + (nxt[j][i-1]==n?0:s[nxt[j][i-1]][i-1]);
      }
    }
  }
  return {nxt, s};
}

int main(){
  ll n, k;scanf("%lld %lld", &n, &k);
  vll a(n);
  re(i, n) scanf("%lld", &a[i]);
  vll b = a;
  reverse(all(b));
  for(int i=0;i<n;i++) b[i] = -b[i];
  vvl nxt, nxt2, s, s2;
  auto tmp = calc(n, k, a);
  nxt = tmp.first;
  s = tmp.second;
  tmp = calc(n, k, b);
  nxt2 = tmp.first;
  s2 = tmp.second;
  auto f = [&](ll l, ll r){//操作回数
    ll cnt = 0;
    ll S = 0;
    ll now = l;
    while(now<r){
      auto idx = upper_bound(all(nxt[now]), r) - nxt[now].begin() - 1;
      if(idx==-1||idx==n||nxt[now][idx]==n||a[nxt[now][idx]]-a[now] < k) break;//もう移動できない
      cnt += (1<<idx);
      S += s[now][idx];
      now = nxt[now][idx];
    }
    return ary{cnt+1, S+l};
  };

  auto g = [&](ll l, ll r){
    l = n - 1 - l, r = n - 1 - r;
    swap(l, r);
    ll cnt = 0;
    ll S = 0;
    ll now = l;
    while(now<r){
      auto idx = upper_bound(all(nxt2[now]), r) - nxt2[now].begin() - 1;
      if(idx==-1||idx==n||nxt2[now][idx]==n||b[nxt2[now][idx]]-b[now]<k) break;//もう移動できない
      cnt += (1<<idx);
      S += s2[now][idx];
      now = nxt2[now][idx];
    }
    return ary{cnt+1, S+l};
  };

  ll q;scanf("%lld", &q);
  for(int i=0;i<q;i++){
    ll x, y;scanf("%lld %lld", &x, &y);
    x--, y--;
    auto h = f(x, y);
    auto u = g(x, y);
    ll cnt = h[0] + 1;
    cnt += u[1] - h[1];
    u[1] = (n-1)*u[0] - u[1];
    ll ret = u[0] + u[1] - h[1];
    printf("%lld\n", ret);
  }
}