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>

#define endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % (k)) + (k) ) % (k))
#define forn(i,n) for (int i = 0; i < n; i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb(x) push_back(x)

using namespace std;

typedef long long ll;
//typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;

const int MX = 200005, LG = 20;
int n, k, a[MX], q;
int pi[MX][LG], pd[MX][LG];
ll si[MX][LG], sd[MX][LG];
int li[MX], ld[MX];

int main () {
	ios_base::sync_with_stdio(0); cin.tie(0);
	
	cin >> n >> k;
	forn (i, n) cin >> a[i];

	for (int i = n - 1; i >= 0; i--) {
		int j = lower_bound(a, a + n, a[i] + k) - a;
		
		if (j == n) {
			pi[i][0] = -1;
			li[i] = 0;
		} else {
			pi[i][0] = j;
			li[i] = li[j] + 1;
		}

		si[i][0] = i;
	}

	for (int i = 0; i < n; i++) {
		int j = upper_bound(a, a + n, a[i] - k) - a;
		j--;

		if (j < 0) {
			pd[i][0] = -1;
			ld[i] = 0;
		} else {
			pd[i][0] = j;
			ld[i] = ld[j] + 1;
		}

		sd[i][0] = i;
	}

	for (int j = 1; j < LG; j++)
		forn (i, n) {
			if (pi[i][j - 1] != -1) {
				pi[i][j] = pi[pi[i][j - 1]][j - 1];
				si[i][j] = si[i][j - 1] + si[pi[i][j - 1]][j - 1];
			} else {
				pi[i][j] = -1;
			}

			if (pd[i][j - 1] != -1) {
				pd[i][j] = pd[pd[i][j - 1]][j - 1];
				sd[i][j] = sd[i][j - 1] + sd[pd[i][j - 1]][j - 1];
			} else {
				pd[i][j] = -1;
			}
		}

	cin >> q;
	while (q--) {
		int l, r;
		cin >> l >> r;
		l--, r--;

		ll sup = 0, to = l, cn = 1;

		for (int j = LG - 1; j >= 0; j--)
			if (pi[to][j] != -1 && pi[to][j] <= r) {
				sup += si[to][j];
				to = pi[to][j];

				cn += 1 << j;
			}

		sup += to;

		ll sdn = 0; to = r;

		for (int j = LG - 1; j >= 0; j--)
			if (pd[to][j] != -1 && pd[to][j] >= l) {
				sdn += sd[to][j];
				to = pd[to][j];
			}

		sdn += to;

		cout << sdn - sup + cn << endl;
	}

	return 0;
}