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;

#define FOR(i, x, y) for(int i = (x); i < (y); ++i)
#define REP(i, x, y) for(int i = (x); i <= (y); ++i)
#define PB push_back
#define MP make_pair
#define PH push
#define fst first
#define snd second
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<int, int> pii;

const int maxn = 2e5 + 5;
const int logn = 20;

int n, k, q;
int a[maxn];
int r[maxn][logn], l[maxn][logn];
ll lsum[maxn][logn], rsum[maxn][logn];

int main(){
	scanf("%d%d", &n, &k);
	FOR(i, 0, n)
		scanf("%d", a + i);
		
	for(int i = 0, j = 0; i < n; ++i){
		for(; j < n && a[j] < a[i] + k; ++j);
		r[i][0] = j;
		rsum[i][0] = j;
	}
	for(int i = n - 1, j = n - 1; i >= 0; --i){
		for(; j >= 0 && a[j] > a[i] - k; --j);
		l[i][0] = j;
		lsum[i][0] = j;
	}
	
	FOR(j, 1, logn) FOR(i, 0, n){
		l[i][j] = l[i][j - 1] == -1 ? -1 : l[l[i][j - 1]][j - 1];
		r[i][j] = r[i][j - 1] == n ? n : r[r[i][j - 1]][j - 1];
		lsum[i][j] = l[i][j - 1] == -1 ? lsum[i][j - 1] : lsum[i][j - 1] + lsum[l[i][j - 1]][j - 1];
		rsum[i][j] = r[i][j - 1] == n ? rsum[i][j - 1] : rsum[i][j - 1] + rsum[r[i][j - 1]][j - 1]; 
	}
	
	scanf("%d", &q);
	FOR(i, 0, q){
		int x, y, len = 0;
		ll sumx = 0, sumy = 0;
		scanf("%d%d", &x, &y); --x; --y;
		int a = x, b = y;
		for(int i = logn - 1; i >= 0; --i){
			if(r[a][i] <= y){
				sumx += rsum[a][i];
				a = r[a][i];
				len += 1 << i;
			}
			if(l[b][i] >= x){
				sumy += lsum[b][i];
				b = l[b][i];
			}
		}
		sumy += y;
		sumx += x;
		++len;
		printf("%lld\n", sumy - sumx + len);
	}
	return 0;
}