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;
typedef long long llo;
#define mp make_pair
#define pb push_back
#define a first 
#define b second
#define endl '\n' 

llo n,k;
llo q;
llo it[200001];
pair<llo,llo> ss[200001][20];
pair<llo,llo> tt[200001][20];
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin>>n>>k;
	for(llo i=0;i<n;i++){
		cin>>it[i];
	}
	for(llo i=n-1;i>=0;i--){
		if(it[n-1]<it[i]+k){
			ss[i][0]={-1,0};
		}
		else{
			llo xx=ss[i+1][0].a;
			if(xx==-1){
				xx=n-1;
			}
			for(llo j=xx;j>=i;j--){
				if(it[j]<it[i]+k){
					break;
				}
				ss[i][0]={j,j};
			}
		}
	}
	for(llo i=0;i<n;i++){
		if(it[0]>it[i]-k){
			tt[i][0]={-1,0};
		}
		else{
			llo xx=tt[i-1][0].a;
			if(xx==-1){
				xx=0;
			}
			for(llo j=xx;j<=i;j++){
				if(it[j]>it[i]-k){
					break;
				}
				tt[i][0]={j,j};
			}
		}
	}
	for(llo j=1;j<20;j++){
		for(llo i=0;i<n;i++){
			if(ss[i][j-1].a==-1){
				ss[i][j]={-1,0};
				continue;
			}
			ss[i][j]={ss[ss[i][j-1].a][j-1].a,-1};
			if(ss[i][j].a==-1){
				continue;
			}
			ss[i][j].b=ss[i][j-1].b+ss[ss[i][j-1].a][j-1].b;
		}
	}
	for(llo j=1;j<20;j++){
		for(llo i=0;i<n;i++){
			if(tt[i][j-1].a==-1){
				tt[i][j]={-1,0};
				continue;
			}
			tt[i][j]={tt[tt[i][j-1].a][j-1].a,-1};
			if(tt[i][j].a==-1){
				continue;
			}
			tt[i][j].b=tt[i][j-1].b+tt[tt[i][j-1].a][j-1].b;
		}
	}
	/*for(llo i=0;i<n;i++){
		cout<<ss[i][0].a<<":";
	}
	cout<<endl;
	for(llo i=0;i<n;i++){
		cout<<tt[i][0].a<<":";
	}
	cout<<endl;
*/

	cin>>q;
	while(q--){
		llo aa,bb;
		cin>>aa>>bb;
		aa--;
		bb--;
		llo x=1;
		llo su=aa;
		llo cur=aa;
		for(llo j=19;j>=0;j--){
			if(ss[cur][j].a==-1){
				continue;
			}
			if(ss[cur][j].a<=bb){
				x+=(1<<j);
				//cout<<j<<":";
				su+=ss[cur][j].b;
				cur=ss[cur][j].a;
			}
		}
		//cout<<endl;
		llo su2=bb;
		cur=bb;
		for(llo j=19;j>=0;j--){
			if(tt[cur][j].a==-1){
				continue;
			}
			if(tt[cur][j].a>=aa){
				su2+=tt[cur][j].b;
				cur=tt[cur][j].a;
			}
		}
		cout<<(su2-su+x)<<endl;
		


	}



	return 0;
}