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;
const int N = 200005;
const int LOGN = 18;
int n, K, a[N], pa1[LOGN][N], pa2[LOGN][N];
long long sum1[LOGN][N], sum2[LOGN][N];
int main() {
	scanf("%d%d", &n, &K);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	for (int i = 1, j = 1; i <= n; i++) {
		while (j <= n && a[j] - a[i] < K) j++;
		pa1[0][i] = j;
		sum1[0][i] = j;
	}
	for (int i = 0; i < LOGN; i++) pa1[i][n + 1] = n + 1;
	for (int i = 1; i < LOGN; i++)
		for (int j = 1; j <= n; j++) {
			pa1[i][j] = pa1[i - 1][pa1[i - 1][j]];
			sum1[i][j] = sum1[i - 1][j] + sum1[i - 1][pa1[i - 1][j]];
		}
	for (int i = n, j = n; i >= 1; i--) {
		while (j > 0 && a[i] - a[j] < K) j--;
		pa2[0][i] = j;
		sum2[0][i] = j;
	}
	for (int i = 0; i < LOGN; i++) pa2[i][0] = 0;
	for (int i = 1; i < LOGN; i++)
		for (int j = 1; j <= n; j++) {
			pa2[i][j] = pa2[i - 1][pa2[i - 1][j]];
			sum2[i][j] = sum2[i - 1][j] + sum2[i - 1][pa2[i - 1][j]];
		}
	int q;
	scanf("%d", &q);
	while (q--) {
		int l, r;
		scanf("%d%d", &l, &r);
		int u = l;
		long long s1 = l;
		int cnt = 1;
		for (int i = LOGN - 1; i >= 0; i--) if (pa1[i][u] <= r) {
			cnt += (1 << i);
			s1 += sum1[i][u];
			u = pa1[i][u];
		}
		u = r;
		long long s2 = r;
		for (int i = LOGN - 1; i >= 0; i--) if (pa2[i][u] >= l) {
			s2 += sum2[i][u];
			u = pa2[i][u];
		}
		printf("%lld\n", s2 - s1 + cnt);
	}
	return 0;
}