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:
/*
读题不规范，爆零两行泪。
数据不清空，爆零两行泪。
多测不读完，爆零两行泪。
边界不特判，爆零两行泪。
贪心不证明，爆零两行泪。
D P 顺序错，爆零两行泪。
大小少等号，爆零两行泪。
变量不统一，爆零两行泪。
越界不判断，爆零两行泪。
调试不注释，爆零两行泪。
溢出不 l l，爆零两行泪。
*/
#include<bits/stdc++.h>
using namespace std;
const int N=200000,LOG_N=20;
int n,m,qu;
int a[N+1];
int rit[N+1],lft[N+1];
int rit_to[N+1][LOG_N],rit_sum[N+1][LOG_N],lft_to[N+1][LOG_N],lft_sum[N+1][LOG_N];
int main(){
	cin>>n>>m;
	for(int i=1;i<=n;i++)scanf("%d",a+i);
	int now=1;
	for(int i=1;i<=n;i++){
		while(now<=n&&a[now]-a[i]<m)now++;
		rit[i]=now;
	}
	rit[n+1]=n+1;
	for(int i=n+1;i;i--){
		rit_to[i][0]=rit[i],rit_sum[i][0]=-rit[i]+1;
		for(int j=1;j<LOG_N;j++)
			rit_to[i][j]=rit_to[rit_to[i][j-1]][j-1],
			rit_sum[i][j]=rit_sum[i][j-1]+rit_sum[rit_to[i][j-1]][j-1];
	}
	now=n;
	for(int i=n;i;i--){
		while(now&&a[i]-a[now]<m)now--;
		lft[i]=now;
	}
	lft[0]=0;
	for(int i=0;i<=n;i++){
		lft_to[i][0]=lft[i],lft_sum[i][0]=lft[i];
		for(int j=1;j<LOG_N;j++)
			lft_to[i][j]=lft_to[lft_to[i][j-1]][j-1],
			lft_sum[i][j]=lft_sum[i][j-1]+lft_sum[lft_to[i][j-1]][j-1];
	}
	cin>>qu;
	while(qu--){
		int l,r;
		scanf("%d%d",&l,&r);
		int ans=0;
		now=l;
		ans+=-now+1;
		for(int i=LOG_N-1;~i;i--)if(rit_to[now][i]<=r)ans+=rit_sum[now][i],now=rit_to[now][i];
		now=r;
		ans+=now;
		for(int i=LOG_N-1;~i;i--)if(lft_to[now][i]>=l)ans+=lft_sum[now][i],now=lft_to[now][i];
		printf("%d\n",ans);
	}
	return 0;
}