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>
#define rep(i,n) for ((i)=1;(i)<=(n);(i)++)
using namespace std;
int n,m,q,i,j,a[200005],l[200005][19],r[200005][19];
long long f[200005][19],g[200005][19];
int main(){
	cin>>n>>m;
	rep(i,n){
		cin>>a[i];
	}
	r[n+1][0]=n+1;
	l[n+1][0]=n+1;
	rep(i,n){
		r[i][0]=max(i,r[i-1][0]);
		while(r[i][0]<=n&&a[r[i][0]]-a[i]<m) r[i][0]++;
	}
	for(i=n;i>=1;i--){
		l[i][0]=min(i,l[i+1][0]);
		while(l[i][0]>=1&&a[i]-a[l[i][0]]<m) l[i][0]--;
	}
	rep(j,18)rep(i,n+1){
		l[i][j]=l[l[i][j-1]][j-1];
		r[i][j]=r[r[i][j-1]][j-1];
	}
	rep(i,n){
		f[i][0]=l[i][0];
		g[i][0]=r[i][0];
	}
	rep(j,18)rep(i,n){
		f[i][j]=f[i][j-1]+f[l[i][j-1]][j-1];
		g[i][j]=g[i][j-1]+g[r[i][j-1]][j-1];
	}
	cin>>q;
	while(q--)
	{
		int x,y,xx,yy;cin>>x>>y;xx=x;yy=y;
		long long ans=y-x+1;
		for(i=18;i>=0;i--){
			if(r[x][i]<=yy){
				ans+=(1<<i);
				ans-=g[x][i];
				x=r[x][i];
			}
			if(l[y][i]>=xx){
				ans+=f[y][i];
				y=l[y][i];
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}