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 maxn=200005;
const int mlog=20;
int n,k,q;
int x[maxn];
int prv[maxn][mlog],nxt[maxn][mlog];
long long sprv[maxn][mlog],snxt[maxn][mlog];

int main(){
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)scanf("%d",&x[i]);
	int ppt=0;
	for(int j=0;j<mlog;j++)prv[0][j]=0;
	for(int i=1;i<=n;i++){
		while(ppt+1<i&&x[i]-x[ppt+1]>=k)ppt++;
		prv[i][0]=ppt;
		sprv[i][0]=i;
		for(int j=1;j<mlog;j++)
			prv[i][j]=prv[prv[i][j-1]][j-1],sprv[i][j]=sprv[i][j-1]+sprv[prv[i][j-1]][j-1];
	}
	int npt=n+1;
	for(int j=0;j<mlog;j++)nxt[n+1][j]=n+1;
	for(int i=n;i>=1;i--){
		while(npt-1>i&&x[npt-1]-x[i]>=k)npt--;
		nxt[i][0]=npt;
		snxt[i][0]=i;
		for(int j=1;j<mlog;j++)
			nxt[i][j]=nxt[nxt[i][j-1]][j-1],snxt[i][j]=snxt[i][j-1]+snxt[nxt[i][j-1]][j-1];
	}
	scanf("%d",&q);
	while(q--){
		int l,r;
		scanf("%d%d",&l,&r);
		long long ans=0;
		int cl=l,cr=r,cd=0;
		for(int i=mlog-1;i>=0;i--)if(nxt[cl][i]<=r){
			ans+=sprv[cr][i]-snxt[cl][i];
			cl=nxt[cl][i];
			cr=prv[cr][i];
			cd+=(1<<i);
		}
		ans+=cr-cl+cd+1;
		printf("%lld\n",ans);
	}
	return 0;
}