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 <cstdio>
#include <cctype>
#include <algorithm>
#include <vector>
#include <cmath>
#include <cassert>
using namespace std;
using ll = long long;

namespace io {
#define File(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)
  const int SIZE = (1 << 21) + 1;
  char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1;
  inline char getc () {return (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++);}
  inline void flush () {fwrite (obuf, 1, oS - obuf, stdout); oS = obuf;}
  inline void putc (char x) {*oS ++ = x; if (oS == oT) flush ();}
  template<class T>
  inline void read(T &x) {
    char ch; int f = 1;
    x = 0;
    while(isspace(ch = getc()));
    if(ch == '-') ch = getc(), f = -1;
    do x = x * 10 + ch - '0'; while(isdigit(ch = getc()));
    x *= f;
  }
  template<class T, class ...Args>
  inline void read(T &x, Args&... args) {read(x); read(args...);}
  template<class T>
  inline void write(T x) {
    static char stk[128];
    int top = 0;
    if(x == 0) {putc('0'); return;}
    if(x < 0) putc('-'), x = -x;
    while(x) stk[top++] = x % 10, x /= 10;
    while(top) putc(stk[--top] + '0');
  }
  template<class T, class ...Args>
  inline void write(T x, Args... args) {write(x); putc(' '); write(args...);}
  inline void space() {putc(' ');}
  inline void endl() {putc('\n');}
  struct _flush {~_flush() {flush();}} __flush;
};
using io::read; using io::write; using io::flush; using io::space; using io::endl; using io::getc; using io::putc;

const int N = 200005;
int lgn;
int n, k;

struct max_good_set {
  int a[N], nxt[N][20];
  ll sum[N][20];
  void init() {
    int p = 0;
    for (int i = 0; i < n; ++i) {
      while (abs(a[p] - a[i]) < k && p < n) ++p;
      nxt[i][0] = p;
      sum[i][0] = i;
    }
    for (int i = 0; i <= lgn; ++i)
      nxt[n][i] = n, sum[n][i] = 0;
    for (int j = 1; j <= lgn; ++j)
      for (int i = 0; i < n; ++i) {
        nxt[i][j] = nxt[nxt[i][j - 1]][j - 1];
        sum[i][j] = sum[i][j - 1] + sum[nxt[i][j - 1]][j - 1];
      }
  }
  pair<ll, int> query(int l, int r) {
    int p = l, c = 1;
    ll s = 0;
    for (int i = lgn; i >= 0; --i)
      if (nxt[p][i] <= r) {
        s += sum[p][i];
        c += 1 << i;
        p = nxt[p][i];
      }
    assert(p <= r);
    return {s + p, c};
  }
}a, b;

int main() {
  read(n, k);
  lgn = log2(n);
  for (int i = 0; i < n; ++i)
    read(a.a[i]);
  reverse_copy(a.a, a.a + n, b.a);
  a.init(); b.init();
  int q;
  read(q);
  while (q--) {
    int l, r;
    read(l, r);
    --l; --r;
    auto A = a.query(l, r), B = b.query(n - 1 - r, n - 1 - l);
    B.first = 1LL * (n - 1) * B.second - B.first;
    write(B.first - A.first + A.second), endl();
  }
  return 0;
}