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 <string>
#include <sstream>
#include <stack>
#include <algorithm>
#include <cmath>
#include <queue>
#include <bitset>
#include <iomanip>
#include <limits>
#include <chrono>
#include <random>
#include <array>
#include <unordered_map>
#include <functional>
#include <complex>
#include <numeric>
#include <cctype>
#include <map>
#include <set>
#include <cstdlib>
#include <bitset>
#include <tuple>
#include <assert.h>
#include <deque>
#include <utility>
#include <fstream>

using namespace std;
typedef long long ll;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T> T gcd(T a, T b) { a = abs(a), b = abs(b); while (b > 0) { tie(a, b) = make_pair(b, a % b); } return a; }
//mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

constexpr long long INF = 1LL << 60;
constexpr int inf = 1000000007;
//constexpr long long mod = 1000000007LL;
//constexpr long long mod = 998244353;
constexpr int MAX = 5000000;


int main()
{

	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	ll n, X; cin >> n >> X;
	const int N = 32;
	vector<ll> x(n); for (int i = 0; i < n; i++) cin >> x[i];
	vector<vector<int>> lfto(N, vector<int>(n + 1, n)), rgto(N, vector<int>(n + 1, -1));
	vector<vector<ll>> lfsm(N, vector<ll>(n + 1)), rgsm(N, vector<ll>(n + 1));
	vector<vector<ll>> lfcn(N, vector<ll>(n + 1)), rgcn(N, vector<ll>(n + 1));
	{
		int cur = 0;
		for (int i = 0; i < n; i++) {
			while (cur < n and x[cur] - x[i] < X) cur++;
			lfto[0][i] = cur, lfsm[0][i] = cur, lfcn[0][i] = 1;
		}
	}
	{
		int cur = n - 1;
		for (int i = n - 1; i >= 0; i--) {
			while (cur >= 0 and x[i] - x[cur] < X) cur--;
			rgto[0][i] = cur, rgsm[0][i] = cur;
		}
	}

	for (int i = 1; i < N; i++) {
		for (int j = 0; j < n; j++) {
			lfto[i][j] = lfto[i - 1][lfto[i - 1][j]];
			if (rgto[i - 1][j] >= 0) rgto[i][j] = rgto[i - 1][rgto[i - 1][j]];
			else rgto[i][j] = -1;
			lfsm[i][j] = lfsm[i - 1][j] + lfsm[i - 1][lfto[i - 1][j]];
			if (rgto[i - 1][j] >= 0) rgsm[i][j] = rgsm[i - 1][j] + rgsm[i - 1][rgto[i - 1][j]];
			lfcn[i][j] = lfcn[i - 1][j] + lfcn[i - 1][lfto[i - 1][j]];
		}
	}
	int kkt; cin >> kkt;
	while (kkt--) {
		int l, r; cin >> l >> r;
		l--; r--;
		ll lf = 0;
		ll rg = 0;
		ll cnt = 1;
		{
			int cur = l;
			for (int i = N - 1; i >= 0; i--) {
				if (lfto[i][cur] <= r) {
					lf += lfsm[i][cur];
					cnt += lfcn[i][cur];
					cur = lfto[i][cur];
				}
			}

		}
		{
			int cur = r;
			for (int i = N - 1; i >= 0; i--) {
				if (rgto[i][cur] >= l) {
					rg += rgsm[i][cur];
					cur = rgto[i][cur];
				}
			}
		}
		lf += l;
		rg += r;
		//cout << lf << " " << rg << " "<< cnt<<endl;
		cout << rg - lf + cnt << "\n";
	}
	return 0;
}