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;

long long n, k, q;
vector<long long> x, rx;
vector<vector<long long>> memo, rmemo, sum, rsum;

int main() {
  cin >> n >> k;
  x.resize(n);
  for (auto &p : x) cin >> p;
  rx = x;
  reverse(rx.begin(), rx.end());
  for (auto &p : rx) p = -p;
  auto calcmemo = [](vector<long long> &v) {
    vector<vector<long long>> res(30, vector<long long>(n + 1, n));
    for (int i = 0; i < n; ++i)
      res[0][i] = lower_bound(v.begin(), v.end(), v[i] + k) - v.begin();
    for (int i = 1; i < 30; ++i)
      for (int j = 0; j < n; ++j) res[i][j] = res[i - 1][res[i - 1][j]];
    return res;
  };
  auto calcsum = [](vector<vector<long long>> &v, bool reversed = 0) {
    vector<vector<long long>> res(30, vector<long long>(n + 1, 0));
    if (reversed)
      for (int i = 1; i <= n; ++i) res[0][i] = v[0][i];
    else
      for (int i = 0; i < n; ++i) res[0][i] = v[0][i] * (v[0][i] != n);
    for (int i = 1; i < 30; ++i)
      for (int j = 0; j <= n; ++j)
        res[i][j] = res[i - 1][j] + res[i - 1][v[i - 1][j]];
    return res;
  };
  memo = calcmemo(x);
  rmemo = calcmemo(rx);
  for (auto &v : rmemo) {
    reverse(v.begin(), v.end());
    for (auto &p : v) p = n - p;
  }
  sum = calcsum(memo);
  rsum = calcsum(rmemo, 1);
  cin >> q;
  for (int i = 0; i < q; ++i) {
    long long x, y, z, now = 0;
    cin >> x >> y;
    --x;
    // x
    z = x;
    for (int j = 29; j >= 0; --j)
      if (memo[j][z] < y) {
        now -= sum[j][z];
        z = memo[j][z];
      }
    // y
    z = y;
    for (int j = 29; j >= 0; --j)
      if (x < rmemo[j][z]) {
        now += rsum[j][z];
        z = rmemo[j][z];
      }
    cout << now + y - x << endl;
  }
  return 0;
}