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 a[200010];
int f1[200010][18], f2[200010][18];
long long s1[200010][18], s2[200010][18];

int main () {
	int n, k; scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	int pos = 1;
	for (int i = 1; i <= n; i++) {
		while (pos <= n && a[pos] - a[i] < k) pos++;
		f1[i][0] = s1[i][0] = pos;
	}
	f1[n + 1][0] = s1[n + 1][0] = n + 1;
	pos = n;
	for (int i = n; i >= 1; i--) {
		while (pos >= 1 && a[i] - a[pos] < k) pos--;
		f2[i][0] = s2[i][0] = pos;
	}
	f2[0][0] = s2[0][0] = 0;
	for (int j = 1; j < 18; j++) {
		for (int i = 0; i <= n + 1; i++) {
			f1[i][j] = f1[f1[i][j - 1]][j - 1];
			s1[i][j] = s1[i][j - 1] + s1[f1[i][j - 1]][j - 1];
			f2[i][j] = f2[f2[i][j - 1]][j - 1];
			s2[i][j] = s2[i][j - 1] + s2[f2[i][j - 1]][j - 1];
		}
	}
	int q; scanf("%d", &q);
	while (q--) {
		int l, r; scanf("%d%d", &l, &r);
		int now = l; long long ss1 = l; int c1 = 1;
		for (int i = 17; i >= 0; i--) {
			if (f1[now][i] <= r) ss1 += s1[now][i], now = f1[now][i], c1 += 1 << i;
		}
		now = r; long long ss2 = r; int c2 = 1;
		for (int i = 17; i >= 0; i--) {
			if (f2[now][i] >= l) ss2 += s2[now][i], now = f2[now][i], c2 += 1 << i;
		}
		assert(c1 == c2);
		printf("%lld\n", ss2 - ss1 + c1);
	}
	return 0;
}