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:

// Problem : D - Keep Distances
// Contest : AtCoder - ACL Contest 1
// URL : https://atcoder.jp/contests/acl1/tasks/acl1_d
// Memory Limit : 1024 MB
// Time Limit : 3000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#pragma GCC optimize("Ofast,-funroll-loops")
#include <bits/stdc++.h>
using namespace std;

constexpr int Maxn = 2e5 + 5;
constexpr int LOG = 20;

int n, K, q;
int a[Maxn];
int pre[Maxn][LOG];
int suf[Maxn][LOG];
int64_t pre_sum[Maxn][LOG];
int64_t suf_sum[Maxn][LOG];

int main() {

  scanf("%d %d", &n, &K);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= n; ++i) {
    pre[i][0] = upper_bound(a + 1, a + n + 1, a[i] - K) - a - 1;
    suf[i][0] = lower_bound(a + 1, a + n + 1, a[i] + K) - a;
    pre_sum[i][0] = (pre[i][0] >= 1 ? pre[i][0] : 0);
    suf_sum[i][0] = (suf[i][0] <= n ? suf[i][0] : 0);
  }
  for (int j = 1; j < LOG; ++j) {
    for (int i = 1; i <= n; ++i) {
      pre_sum[i][j] = pre_sum[i][j - 1];
      suf_sum[i][j] = suf_sum[i][j - 1];
      if (pre[i][j - 1] >= 1) {
        pre[i][j] = pre[pre[i][j - 1]][j - 1];
        pre_sum[i][j] += pre_sum[pre[i][j - 1]][j - 1];
      }
      else {
        pre[i][j] = 0;
      }
      if (suf[i][j - 1] <= n) {
        suf[i][j] = suf[suf[i][j - 1]][j - 1];
        suf_sum[i][j] += suf_sum[suf[i][j - 1]][j - 1];
      }
      else {
        suf[i][j] = n + 1;
      }
    }
  }
  
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int64_t left_sum = 0;
    int64_t right_sum = 0;
    int len = 1;
    {
      int lg = LOG - 1, pos = r;
      while (lg >= 0) {
        if (pre[pos][lg] >= l) {
          left_sum += pre_sum[pos][lg];
          pos = pre[pos][lg];
          len += (1 << lg);
        }
        --lg;
      }
    }
    {
      int lg = LOG - 1, pos = l;
      while (lg >= 0) {
        if (suf[pos][lg] <= r) {
          right_sum += suf_sum[pos][lg];
          pos = suf[pos][lg];
        }
        --lg;
      }
    }
    printf("%lld\n", left_sum - right_sum + len + r - l);
  }
  
  return 0;
  
}