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 <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#define llint long long
#define inf 1e18
#define rep(x, s, t) for(llint (x) = (s); (x) < (t); (x)++)
#define Rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))

using namespace std;
typedef pair<llint, llint> P;

llint n, k, Q;
llint x[200005];
llint pred[19][200005], succ[19][200005];
llint pred2[19][200005], succ2[19][200005];

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> k;
	for(int i = 1; i <= n; i++) cin >> x[i];
	x[0] = -inf, x[n+1] = inf;
	
	for(int i = 1; i <= n; i++){
		llint p = lower_bound(x, x+n+2, x[i]+k) - x;
		succ[0][i] = p;
		p = upper_bound(x, x+n+2, x[i]-k) - x - 1;
		pred[0][i] = p;
	}
	succ[0][n+1] = n+1;
	
	for(int i = 1; i < 19; i++){
		for(int j = 1; j <= n+1; j++){
			pred[i][j] = pred[i-1][pred[i-1][j]];
			succ[i][j] = succ[i-1][succ[i-1][j]];
		}
	}
	
	for(int i = 1; i <= n; i++){
		succ2[0][i] = succ[0][i]-1;
		pred2[0][i] = pred[0][i];
	}
	for(int i = 1; i < 19; i++){
		for(int j = 1; j <= n; j++){
			pred2[i][j] = pred2[i-1][j] + pred2[i-1][pred[i-1][j]];
			succ2[i][j] = succ2[i-1][j] + succ2[i-1][succ[i-1][j]];
		}
	}
	
	/*for(int i = 1; i <= n; i++) cout << succ[0][i] << " "; cout << endl;
	for(int i = 1; i <= n; i++) cout << pred[0][i] << " "; cout << endl;
	
	for(int i = 1; i <= n; i++) cout << succ2[0][i] << " "; cout << endl;
	for(int i = 1; i <= n; i++) cout << pred2[0][i] << " "; cout << endl;*/
	
	cin >> Q;
	llint L, R;
	for(int q = 1; q <= Q; q++){
		cin >> L >> R;
		
		llint ans = R-L+1, v = L;
		for(int i = 18; i >= 0; i--){
			//cout << succ[i][v] << " " << R << endl;
			if(succ[i][v] <= R){
				ans -= succ2[i][v];
				v = succ[i][v];
			}
			///cout << v << " " << ans << endl;
		}
		v = R;
		for(int i = 18; i >= 0; i--){
			if(pred[i][v] >= L){
				ans += pred2[i][v];
				v = pred[i][v];
			}
		}
		cout << ans << endl;
	}
	
	return 0;
}