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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__);fflush(stderr);
#else
	#define eprintf(...) 42
#endif

using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
template<typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll myRand(ll B) {
	return (ull)rng() % B;
}

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

clock_t startTime;
double getCurrentTime() {
	return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}

const int N = 200200;
const int LOG = 18;
int a[N];
int n, d;
ll b[LOG][N][2];
ll c[LOG][N][2];

ll solveL(int l, int r) {
	ll ans = 0;
	int len = 0;
	for (int k = LOG - 1; k >= 0; k--) {
		if (b[k][l][0] <= r) {
			len += 1 << k;
			ans += b[k][l][1];
			l = b[k][l][0];
		}
	}
//	eprintf("len = %d\n", len + 1);
	return ans + l;
}
ll solveR(int l, int r) {
	ll ans = 0;
	int len = 0;
	for (int k = LOG - 1; k >= 0; k--) {
		if (c[k][r][0] >= l) {
			len += 1 << k;
			ans += c[k][r][1];
			r = c[k][r][0];
		}
	}
//	eprintf("len = %d\n", len + 1);
	return ans + r + 1;
}

int main()
{
	startTime = clock();
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	scanf("%d%d", &n, &d);
	for (int i = 1; i <= n; i++)
		scanf("%d", &a[i]);

	int p = 1;
	for (int i = 1; i <= n; i++) {
		while(p <= n && a[p] - a[i] < d) p++;
		//eprintf("%d %d\n", i, p);
		b[0][i][0] = p;
		b[0][i][1] = i;
	}
	b[0][n + 1][0] = n + 1;

	for (int k = 0; k < LOG - 1; k++) {
		for (int i = 1; i <= n + 1; i++) {
			int v = b[k][i][0];
			b[k + 1][i][0] = b[k][v][0];
			b[k + 1][i][1] = b[k][i][1] + b[k][v][1];
		}
	}

	p = n;
	for (int i = n; i >= 1; i--) {
		while(p > 0 && a[i] - a[p] < d) p--;
		//eprintf("%d %d\n", i, p);
		c[0][i][0] = p;
		c[0][i][1] = i + 1;
	}
	c[0][0][0] = 0;

	for (int k = 0; k < LOG - 1; k++) {
		for (int i = 0; i <= n; i++) {
			int v = c[k][i][0];
			c[k + 1][i][0] = c[k][v][0];
			c[k + 1][i][1] = c[k][i][1] + c[k][v][1];
		}
	}

	int q;
	scanf("%d", &q);
	while(q--) {
		int l, r;
		scanf("%d%d", &l, &r);
		printf("%lld\n", solveR(l, r) - solveL(l, r));
	}

	return 0;
}