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;
const int N=200005;
int n,k,fa[20][N],q,i,l[N],r[N],x[N],j;
long long ans[N],faf[20][N];
int main()
{
	scanf("%d %d",&n,&k);
	for(i=1;i<=n;++i)
		scanf("%d",&x[i]);
	scanf("%d",&q);
	for(i=1;i<=q;++i)
		scanf("%d %d",&l[i],&r[i]);
	x[0]=-k-1;
	x[n+1]=x[n]+k+1;
	for(i=n;i>=1;--i)
	{
		fa[0][i]=upper_bound(x+1,x+1+n,x[i]-k)-x-1;
		faf[0][i]=i;
	}
	for(i=1;i<20;++i)
		for(j=1;j<=n;++j)
		{
			fa[i][j]=fa[i-1][fa[i-1][j]];
			faf[i][j]=faf[i-1][j]+faf[i-1][fa[i-1][j]];
		}
	for(i=1;i<=q;++i)
	{
		int p=r[i];
		for(j=19;j>=0;--j)
			if(fa[j][p]>=l[i])
			{
				ans[i]+=faf[j][p];
				p=fa[j][p];
			}
		ans[i]+=faf[0][p];
	}

	for(i=1;i<=n;++i)
	{
		fa[0][i]=lower_bound(x+1,x+2+n,x[i]+k)-x;
		faf[0][i]=i-1;
	}
	fa[0][n+1]=n+1;
	for(i=1;i<20;++i)
		for(j=1;j<=n+1;++j)
		{
			fa[i][j]=fa[i-1][fa[i-1][j]];
			faf[i][j]=faf[i-1][j]+faf[i-1][fa[i-1][j]];
		}
	for(i=1;i<=q;++i)
	{
		int p=l[i];
		for(j=19;j>=0;--j)
			if(fa[j][p]<=r[i])
			{
				ans[i]-=faf[j][p];
				p=fa[j][p];
			}
		ans[i]-=faf[0][p];
	}
	for(i=1;i<=q;++i)
		printf("%lld\n",ans[i]);
}