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>
using namespace std;

#define INF 1234567890
#define ll long long

int N, K, Q;
int X[200201];
int L[20][200201], R[20][200201]; //
ll LS[20][200201], RS[20][200201]; //
// LS[i][j] : j를 포함해서 만나는 2^i개의 정점의 번호 합

int main()
{
	scanf("%d %d", &N, &K);
	for(int i=1; i<=N; i++)
		scanf("%d", &X[i]);

	// L[] 채우기
	map<int, int> m;
	for(int i=1; i<=N; i++)
	{
		m[X[i]] = i;
		auto it = m.upper_bound(X[i]-K);
		if (it == m.begin()) continue;
		
		it = prev(it);
		L[0][i] = it->second;
	}

	m.clear();
	for(int i=N; i>=1; i--)
	{
		m[X[i]] = i;
		auto it = m.lower_bound(X[i]+K);
		if (it == m.end()) continue;

		R[0][i] = it->second;
	}

	for(int i=1; i<=N; i++)
		LS[0][i] = RS[0][i] = i;

	for(int i=1; i<20; i++)
		for(int j=1; j<=N; j++)
		{
			L[i][j] = L[i-1][L[i-1][j]];
			R[i][j] = R[i-1][R[i-1][j]];
			LS[i][j] = LS[i-1][j] + LS[i-1][L[i-1][j]];
			RS[i][j] = RS[i-1][j] + RS[i-1][R[i-1][j]];
		}

	scanf("%d", &Q);
	while(Q--)
	{
		int l, r;
		scanf("%d %d", &l, &r);
		ll ls = 0, rs = 0, cnt = 0;

		int n = r;
		for(int i=19; i>=0; i--)
			if (L[i][n] != 0 && l <= L[i][n])
				rs += LS[i][n], n = L[i][n], cnt += (1<<i);
		rs += LS[0][n]; cnt++;

		n = l;
		for(int i=19; i>=0; i--)
			if (R[i][n] != 0 && R[i][n] <= r)
				ls += RS[i][n], n = R[i][n];
		ls += RS[0][n];

		//printf("%lld, %lld, %lld : ", rs, ls, cnt);
		printf("%lld\n", rs-ls+cnt);
	}
	return 0;
}