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 ll long long
const int _ = 2e5 + 7; int tor[_][20] , tol[_][20]; ll sumr[_][20] , suml[_][20]; int N , K , Q , X[_];

int main(){
	ios::sync_with_stdio(0); cin >> N >> K; for(int i = 1 ; i <= N ; ++i) cin >> X[i];
	X[0] = -1e9; X[N + 1] = 2e9; for(int j = 0 ; j < 20 ; ++j) tor[N + 1][j] = N + 1;
	for(int pos = N + 1 , i = N ; i ; --i){
		while(X[pos - 1] - X[i] >= K){--pos;} tor[i][0] = pos; sumr[i][0] = pos - 1;
		for(int j = 1 ; j < 20 ; ++j){tor[i][j] = tor[tor[i][j - 1]][j - 1]; sumr[i][j] = sumr[i][j - 1] + sumr[tor[i][j - 1]][j - 1];}
	}
	for(int pos = 0 , i = 1 ; i <= N ; ++i){
		while(X[i] - X[pos + 1] >= K){++pos;} tol[i][0] = pos; suml[i][0] = pos;
		for(int j = 1 ; j < 20 ; ++j){tol[i][j] = tol[tol[i][j - 1]][j - 1]; suml[i][j] = suml[i][j - 1] + suml[tol[i][j - 1]][j - 1];}
	}

	for(cin >> Q ; Q ; --Q){
		int L , R; cin >> L >> R; ll ans = R - L + 1;
		for(int t = R , fl = 19 ; ~fl ; --fl) if(tol[t][fl] >= L){ans += suml[t][fl]; t = tol[t][fl];}
		for(int t = L , fl = 19 ; ~fl ; --fl) if(tor[t][fl] <= R){ans -= sumr[t][fl]; t = tor[t][fl];}
		cout << ans << endl;
	}
	return 0;
}