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>

using namespace std;

int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  vector<int> x(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &x[i]);
  }
  vector<vector<int>> to_lef(32 - __builtin_clz(n)), to_rig(32 - __builtin_clz(n));
  vector<vector<long long>> sum_lef(32 - __builtin_clz(n)), sum_rig(32 - __builtin_clz(n));
  for (int i = 0, j = -1; i < n; ++i) {
    while (j + 1 < n && x[i] - x[j + 1] >= k) ++j;
    to_lef[0].push_back(j); 
    sum_lef[0].push_back(i);
  }
  for (int i = 0, j = 0; i < n; ++i) {
    while (j < n && x[j] - x[i] < k) ++j;
    to_rig[0].push_back(j);
    sum_rig[0].push_back(i-1);
  }
  for (int i = 0; i + 1 < (int)to_lef.size(); ++i) {
    to_lef[i+1].resize(n);
    sum_lef[i+1].resize(n);
    for (int j = 0; j < n; ++j) {
      int now = to_lef[i][j];
      long long sum = sum_lef[i][j];
      if (now != -1) {
        sum += sum_lef[i][now];
        now = to_lef[i][now];
      }
      to_lef[i+1][j] = now;
      sum_lef[i+1][j] = sum;
    }
  }
  for (int i = 0; i + 1 < (int)to_rig.size(); ++i) {
    to_rig[i+1].resize(n);
    sum_rig[i+1].resize(n);
    for (int j = 0; j < n; ++j) {
      int now = to_rig[i][j];
      long long sum = sum_rig[i][j];
      if (now < n) {
        sum += sum_rig[i][now];
        now = to_rig[i][now];
      }
      to_rig[i+1][j] = now;
      sum_rig[i+1][j] = sum;
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    --l, --r;
    long long ans = 0;
    int now = r;
    for (int i = (int) to_lef.size()-1; i >= 0; --i) {
      if (to_lef[i][now] >= l) {
        ans += sum_lef[i][now];
        now = to_lef[i][now];
      }
    }
    ans += now;
    now = l;
    for (int i = (int) to_rig.size()-1; i >= 0; --i) {
      if (to_rig[i][now] <= r) {
        ans -= sum_rig[i][now];
        now = to_rig[i][now];
      }
    }
    ans -= now - 1;
    printf("%lld\n", ans);
  }
  return 0;
}