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 <stdio.h>
#include <queue>
using namespace std;

typedef long long int ll;
constexpr int kN = int(2E5 + 10), LOGN = 20, kInf = int(2E9 + 10);

int x[kN], nxtl[kN][LOGN], nxtr[kN][LOGN];
ll pl[kN][LOGN], pr[kN][LOGN];

int main() {
	int n, k, q, l, r, nl, nr;
	ll ans;
	queue<int> qu;
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
	x[0] = -kInf, x[++n] = kInf;
	for (int i = 0; i <= n; i++) {
		while (!qu.empty()) {
			if (x[i] - x[qu.front()] >= k) {
				nxtl[qu.front()][0] = i;
				qu.pop();
			}
			else break;
		}
		qu.push(i);
	}
	while (!qu.empty()) {
		nxtl[qu.front()][0] = n;
		qu.pop();
	}

	for (int i = n; i >= 0; i--) {
		while (!qu.empty()) {
			if (x[qu.front()] - x[i] >= k) {
				nxtr[qu.front()][0] = i;
				qu.pop();
			}
			else break;
		}
		qu.push(i);
	}
	while (!qu.empty()) {
		nxtr[qu.front()][0] = 0;
		qu.pop();
	}

	for (int i = 1; i < LOGN; i++) for (int j = 0; j <= n; j++) nxtl[j][i] = nxtl[nxtl[j][i - 1]][i - 1];
	for (int i = 1; i < LOGN; i++) for (int j = 0; j <= n; j++) nxtr[j][i] = nxtr[nxtr[j][i - 1]][i - 1];
	for (int i = 0; i <= n; i++) pl[i][0] = nxtl[i][0];
	for (int i = 0; i <= n; i++) pr[i][0] = nxtr[i][0];
	for (int i = 0; i <= n; i++) if (pl[i][0] == n) pl[i][0] = 0;


	for (int i = 1; i < LOGN; i++) for (int j = 0; j <= n; j++) pl[j][i] = pl[j][i - 1] + pl[nxtl[j][i - 1]][i - 1];
	for (int i = 1; i < LOGN; i++) for (int j = 0; j <= n; j++) pr[j][i] = pr[j][i - 1] + pr[nxtr[j][i - 1]][i - 1];
	scanf("%d", &q);
	for (int i = 1; i <= q; i++) {
		scanf("%d%d", &l, &r);
		nl = l, nr = r;
		ans = r - l + 1;
		for (int j = LOGN - 1; j >= 0; j--) {
			if (nxtl[l][j] <= r) {
				ans -= pl[l][j];
				l = nxtl[l][j];
				ans += 1 << j;
			}
		}
		l = nl, r = nr;
		for (int j = LOGN - 1; j >= 0; j--) {
			if (nxtr[r][j] >= l) {
				ans += pr[r][j];
				r = nxtr[r][j];
			}
		}
		printf("%lld\n", ans);
	}
}