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<iostream>
using namespace std;
int N,K;
int X[2<<17];
int dL[20][2<<17],dLs[20][2<<17];
int dR[20][2<<17],dRs[20][2<<17];
int Q;
int main()
{
	cin>>N>>K;
	for(int i=0;i<N;i++)cin>>X[i];
	int l=0;
	for(int i=0;i<N;i++)
	{
		while(l<N&&X[l]<X[i]+K)l++;
		dL[0][i]=l;
		dLs[0][i]=i;
	}
	int r=N-1;
	for(int i=N;i--;)
	{
		while(r>=0&&X[r]>X[i]-K)r--;
		dR[0][i]=r;
		dRs[0][i]=i;
	}
	for(int k=0;k<19;k++)
	{
		for(int i=0;i<N;i++)
		{
			if(dL[k][i]==N)
			{
				dL[k+1][i]=N;
				dLs[k+1][i]=dLs[k][i];
			}
			else
			{
				dL[k+1][i]=dL[k][dL[k][i]];
				dLs[k+1][i]=dLs[k][i]+dLs[k][dL[k][i]];
			}
			if(dR[k][i]==-1)
			{
				dR[k+1][i]=-1;
				dRs[k+1][i]=dRs[k][i];
			}
			else
			{
				dR[k+1][i]=dR[k][dR[k][i]];
				dRs[k+1][i]=dRs[k][i]+dRs[k][dR[k][i]];
			}
		}
	}
	cin>>Q;
	for(;Q--;)
	{
		int L,R;cin>>L>>R;
		L--;
		int cnt=1,now=L;
		int ans=0;
		for(int k=20;k--;)
		{
			if(dL[k][now]<R)
			{
				cnt+=1<<k;
				ans-=dLs[k][now];
				now=dL[k][now];
			}
		}
		ans-=now;
		now=R-1;
		for(int k=20;k--;)
		{
			if(dR[k][now]>=L)
			{
				ans+=dRs[k][now];
				now=dR[k][now];
			}
		}
		ans+=now;
		cout<<ans+cnt<<endl;
	}
}