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 <bits/stdc++.h>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define show(x) cerr<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef tuple<int,int,int> T;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
typedef vector<T> vt;
int getInt(){int x;scanf("%d",&x);return x;}
template<typename T>istream& operator>>(istream&i,v(T)&v){rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const v(T)&v){stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>ostream& operator<<(ostream&o,const v(T)&v){if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>istream& operator>>(istream&i,pair<T1,T2>&v){return i>>v.fi>>v.se;}
template<typename T1,typename T2>ostream& operator<<(ostream&o,const pair<T1,T2>&v){return o<<v.fi<<","<<v.se;}
template<typename T>bool mins(T& x,const T&y){if(x>y){x=y;return true;}else return false;}
template<typename T>bool maxs(T& x,const T&y){if(x<y){x=y;return true;}else return false;}
template<typename T>ll suma(const v(T)&a){ll res(0);for(auto&&x:a)res+=x;return res;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#define dame { puts("-1"); return 0;}
#define yn {puts("Yes");}else{puts("No");}
const int MX = 200005;
const int D = 18;
typedef v(vl) vvl;


int n,K;

struct Db {
  vvi d; vvl s;
  Db(vi x):d(D,vi(n+1,n)),s(D,vl(n+1,0)) {
    int j = n;
    drep(i,n) {
      while (x[i]+K <= x[j-1]) --j;
      d[0][i] = j;
      s[0][i] = j;
    }
    rep(i,D-1) {
      rep(j,n) {
        d[i+1][j] = d[i][d[i][j]];
        s[i+1][j] = s[i][j]+s[i][d[i][j]];
      }
    }
  }
  int cnt;
  ll get(int l, int r) {
    int j = l;
    ll res = l;
    cnt = 1;
    int w = 1<<D;
    drep(i,D) {
      w >>= 1;
      if (d[i][j] > r) continue;
      res += s[i][j];
      j = d[i][j];
      cnt += w;
    }
    return res;
  }
};

int main() {
  scanf("%d%d",&n,&K);
  vi x(n);
  cin>>x;

  Db dl(x);
  reverse(rng(x));
  rep(i,n) x[i] = -x[i];
  Db dr(x);

  int q;
  scanf("%d",&q);
  rep(qi,q) {
    int l,r;
    scanf("%d%d",&l,&r);
    --l; --r;
    ll xr = dr.get(n-1-r,n-1-l);
    xr = dr.cnt*ll(n-1) - xr;
    ll ans = xr - dl.get(l,r) + dl.cnt;
    printf("%lld\n",ans);
  }
  return 0;
}