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(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;

int n, k;
int l, r;
int a[200200];
int pre_pos[22][200200], suf_pos[22][200200];
int q;
ll pre_sum[22][200200], suf_sum[22][200200];

int main(){
	ios::sync_with_stdio(false);
	cin >> n >> k;
	rep(i, n) cin >> a[i];
	r = 0;
	for(int i = 0; i < n; i++){
		while(r < n && a[r] - a[i] < k) r++;
		suf_pos[0][i] = r;
		suf_sum[0][i] = i;
	}
	l = n-1;
	for(int i = n-1; i >= 0; i--){
		while(l >= 0 && a[i] - a[l] < k) l--;
		pre_pos[0][i] = l;
		pre_sum[0][i] = i;
	}
	for(int i = 0; i < 20; i++){
		for(int j = 0; j < n; j++){
			pre_pos[i + 1][j] = (pre_pos[i][j] < 0) ? -1 : pre_pos[i][pre_pos[i][j]];
			pre_sum[i + 1][j] = (pre_pos[i][j] < 0) ? pre_sum[i][j] : (ll)(pre_sum[i][j] + pre_sum[i][pre_pos[i][j]]);
			suf_pos[i + 1][j] = (suf_pos[i][j] >= n) ? n : suf_pos[i][suf_pos[i][j]];
			suf_sum[i + 1][j] = (suf_pos[i][j] >= n) ? suf_sum[i][j] : (ll)(suf_sum[i][j] + suf_sum[i][suf_pos[i][j]]);
		}
	}
	cin >> q;
	while(q--){
		cin >> l >> r;
		l--, r--;
		int nl = l, nr = r;
		int l_ans = 1, r_ans = 1; 
		ll l_sum = 0, r_sum = 0;
		for(int i = 20; i >= 0; i--){
			if(suf_pos[i][nl] <= r){
				l_ans += (1 << i);
				l_sum += suf_sum[i][nl];
				nl = suf_pos[i][nl];
			}
		}
		l_sum += nl;
		for(int i = 20; i >= 0; i--){
			if(pre_pos[i][nr] >= l){
				r_ans += (1 << i);
				r_sum += pre_sum[i][nr];
				nr = pre_pos[i][nr];
			}
		}
		r_sum += nr;
		assert(l_ans == r_ans);
		cout << r_sum - l_sum + l_ans << endl;
	}
	return 0;
}