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 ll;
typedef unsigned long long ull;
typedef pair<int,int> ii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
#define F first
#define S second
#define RE register
#define random(x) (rand()%x)
#define LOG(a,b) (log(a)/log(b))
#define N 200005
#define mod 1000000007
#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3fll
int n,lgn,q,k,x[N],lp[22][N],rp[22][N];
ll lsz[22][N],rsz[22][N];
int main(){
	scanf("%d%d",&n,&k);
	lgn=log2(n);
	for(int i=1;i<=n;++i){
		scanf("%d",x+i);
	}
	for(int i=0;i<=lgn;++i) rp[i][n+1]=n+1;
	for(int i=1;i<=n;++i){
		lp[0][i]=lower_bound(x+1,x+n+1,x[i]-k+1)-x-1;
		lsz[0][i]=lp[0][i];
		//printf("(%d:%d,%lld) ",0,lp[0][i],lsz[0][i]);
		for(int j=1;j<=lgn;++j){
			lp[j][i]=lp[j-1][lp[j-1][i]];
			lsz[j][i]=lsz[j-1][i]+lsz[j-1][lp[j-1][i]];
			//printf("(%d:%d,%lld) ",j,lp[j][i],lsz[j][i]);
		}
		//puts("");
	}
	for(int i=n;i;--i){
		rp[0][i]=lower_bound(x+1,x+n+1,x[i]+k)-x;
		rsz[0][i]=rp[0][i];
		for(int j=1;j<=lgn;++j){
			rp[j][i]=rp[j-1][rp[j-1][i]];
			rsz[j][i]=rsz[j-1][i]+rsz[j-1][rp[j-1][i]];
			//printf("(%d:%d,%lld) ",j,rp[j][i],rsz[j][i]);
		}
		//puts("");
	}
	scanf("%d",&q);
	int ql,qr,p,cntl=1,cntr=1;
	ll ansl=0,ansr=0;
	for(int i=1;i<=q;++i){
		scanf("%d%d",&ql,&qr);
		p=ql;ansl=ql;cntl=1;
		for(int j=lgn;j>=0;--j){
			if(rp[j][p]<=qr){
				cntl+=(1<<j);
				ansl+=rsz[j][p];
				p=rp[j][p];
			}
			//printf("%d:%d %d\n",j,p,ansl);
		}
		p=qr;ansr=qr;cntr=1;
		for(int j=lgn;j>=0;--j){
			if(lp[j][p]>=ql){
				cntr+=(1<<j);
				ansr+=lsz[j][p];
				p=lp[j][p];
			}
			//printf("%d:%d %d\n",j,p,ansr);
		}
		//printf("%d %d\n",cntl,cntr);
		assert(cntl==cntr);
		printf("%lld\n",ansr-ansl+cntl);
	}
	return 0;
}