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 "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "unordered_map"
#include "unordered_set"
#include "iomanip"
#include "cmath"
#include "random"
#include "bitset"
#include "cstdio"
#include "numeric"
#include "cassert"
#include "ctime"

using namespace std;

//constexpr long long int MOD = 1000000007;
//constexpr int MOD = 1000000007;
//constexpr int MOD = 998244353;
constexpr long long int MOD = 998244353;
constexpr double EPS = 1e-12;

//int N, M, K, T, H, W, L, R;
long long int N, M, K, T, H, W, L, R;


int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin >> N >> K;
	vector<long long int>v(N + 2);
	for (int i = 1; i <= N; i++)cin >> v[i];
	v[0] = -MOD * MOD;
	v.back() = MOD * MOD;
	cin >> T;
	N += 2;
	vector<vector<long long int>>ord_tapi(N, vector<long long int>(20, -1));
	vector<vector<long long int>>ord_sum(N, vector<long long int>(20));
	vector<vector<long long int>>rev_tapi(N, vector<long long int>(20, -1));
	vector<vector<long long int>>rev_sum(N, vector<long long int>(20));
	int cnt = 0;
	for (int i = 0; i < N; i++) {
		while (cnt != -1 && v[cnt] - v[i] < K) {
			cnt++;
			if (cnt == N) {
				cnt = -1;
				break;
			}
		}
		ord_tapi[i][0] = cnt;
		ord_sum[i][0] = i;
	}
	for (int j = 1; j < 20; j++) {
		for (int i = 0; i < N; i++) {
			if (ord_tapi[i][j - 1] == -1)continue;
			ord_tapi[i][j] = ord_tapi[ord_tapi[i][j - 1]][j - 1];
			ord_sum[i][j] = ord_sum[i][j - 1] + ord_sum[ord_tapi[i][j - 1]][j - 1];
		}
	}
	cnt = N - 1;
	for (int i = N - 1; i >= 0; i--) {
		while (cnt != -1 && v[i] - v[cnt] < K) {
			cnt--;
			if (cnt == -1) {
				cnt = -1;
				break;
			}
		}
		rev_tapi[i][0] = cnt;
		rev_sum[i][0] = i;
	}
	for (int j = 1; j < 20; j++) {
		for (int i = 0; i < N; i++) {
			if (rev_tapi[i][j - 1] == -1)continue;
			rev_tapi[i][j] = rev_tapi[rev_tapi[i][j - 1]][j - 1];
			rev_sum[i][j] = rev_sum[i][j - 1] + rev_sum[rev_tapi[i][j - 1]][j - 1];
		}
	}
	while (T--) {
		cin >> L >> R;
		long long int ans = 0;
		int num = 0;
		int p = L;
		for (int cnt = 19; cnt >= 0; cnt--) {
			if ((ord_tapi[p][cnt] != -1 && ord_tapi[p][cnt] <= R)) {
				num |= 1 << cnt;
				ans -= ord_sum[p][cnt];
				p = ord_tapi[p][cnt];
			}
		}
		ans -= ord_sum[p][0];
	//	cout << num << endl;
	//	cout << ans << endl;
		num = 0;
		p = R;
		for (int cnt = 19; cnt >= 0; cnt--) {
			if ((rev_tapi[p][cnt] != -1 && rev_tapi[p][cnt] >= L)) {
				num |= 1 << cnt;
				ans += rev_sum[p][cnt];
				p = rev_tapi[p][cnt];
			}
		}
		ans += rev_sum[p][0];
		cout << ans + num + 1 << endl;
	}
}