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, M = 21, inf = (int)2e9;

int n, k, q, x[N];
int fa1[N][M], fa2[N][M], Log2[N];
long long val1[N][M], val2[N][M];
int main() {
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; i++) scanf("%d", &x[i]);

	x[0] = -inf, x[n + 1] = inf;
	fa1[n + 1][0] = n + 1, fa2[0][0] = 0;
	val1[n + 1][0] = 0, val2[0][0] = 0;
	Log2[1] = 0;
	for (int i = 2; i <= n; i++) Log2[i] = Log2[i >> 1] + 1;
	for (int i = 1; i <= n; i++) {
		fa1[i][0] = lower_bound(x + 1, x + n + 1, x[i] + k) - x;
		fa2[i][0] = upper_bound(x + 1, x + n + 1, x[i] - k) - x - 1;
		val1[i][0] = i, val2[i][0] = i + 1;
	}
	for (int i = 1; i <= Log2[n]; i++) {
		for (int j = 1; j <= n + 1; j++) {
			fa1[j][i] = fa1[fa1[j][i - 1]][i - 1];
			val1[j][i] = val1[j][i - 1] + val1[fa1[j][i - 1]][i - 1];
		}
		for (int j = 0; j <= n; j++) {
			fa2[j][i] = fa2[fa2[j][i - 1]][i - 1];
			val2[j][i] = val2[j][i - 1] + val2[fa2[j][i - 1]][i - 1];
		}
	}

	scanf("%d", &q);
	for (int i = 1; i <= q; i++) {
		int l, r, L, R;
		long long ans = 0;
		scanf("%d%d", &l, &r);
		L = l, R = r;
		for (int j = Log2[n - l + 1]; j >= 0; j--) {
			if (fa1[L][j] <= r) ans -= val1[L][j], L = fa1[L][j];
		}
		ans -= val1[L][0];
		for (int j = Log2[r]; j >= 0; j--) {
			if (fa2[R][j] >= l) ans += val2[R][j], R = fa2[R][j];
		}
		ans += val2[R][0];
		printf("%lld\n", ans);
	}
	return 0;
}