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>
#ifdef BIZON
	#define rr(x) cerr << "\e[1;38;5;172m" << #x << "\e[0m = " << (x) << endl;
#else
	#define rr(x)
	#define endl '\n'
#endif
#define ALL(c) begin(c), end(c)
using namespace std;
using ll = long long;
using ld = long double;

int __; 
#define in (cin>>__, (int)__)


int main(){
	if(auto f="in.txt"; fopen(f,"r") && freopen(f,"r",stdin));
	ios::sync_with_stdio(0);cin.tie(0);//cout.precision(9);cout<<fixed;
	
	
	int n = in, D = in;
	
	vector<int> x(n);
	for(auto &_ : x) cin>>_;
	
	constexpr int L = 18;
	vector<array<pair<int,ll>,L>> pl(n), pr(n);
	
	for(int i=n-1, j = n; i>=0; --i){
		while(j-1>i && abs(x[j-1]-x[i])>=D) --j;
		pr[i][0] = {j, 0};
		for(int l=1;l<L;++l){
			int k = pr[i][l-1].first;
			if(k==n) pr[i][l] = pr[i][l-1];
			else pr[i][l] = {pr[k][l-1].first, pr[k][l-1].second + pr[i][l-1].second + k};
		}
	}
	
	for(int i=0, j = -1; i<n; ++i){
		while(j+1<i && abs(x[j+1]-x[i])>=D) ++j;
		pl[i][0] = {j, 0};
		for(int l=1;l<L;++l){
			int k = pl[i][l-1].first;
			if(k==-1) pl[i][l] = pl[i][l-1];
			else pl[i][l] = {pl[k][l-1].first, pl[k][l-1].second + pl[i][l-1].second + k};
		}
	}
	
	
	int m = in;
	while(m--){
		int l = in-1, r = in-1;
		
		int fr = l, cr = 1;
		ll sr = 0;
		for(int k=L-1; k>=0; --k) if(pr[fr][k].first<=r) 
			sr+=fr+pr[fr][k].second, fr = pr[fr][k].first, cr+=1<<k;
		sr+=fr;
		
		
		int fl = r, cl = 1;
		ll sl = 0;
		for(int k=L-1; k>=0; --k) if(pl[fl][k].first>=l) 
			sl+=fl+pl[fl][k].second, fl = pl[fl][k].first, cl+=1<<k;
		sl+=fl;
		
		//cout<<cl<<' '<<cr<<endl;
		assert(cl==cr);
		
		ll res = cl + sl - sr;
		cout<<res<<endl;
	}
	
	return 0;
}