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 fi first
#define se second

const int N = 200200;
const int mod = 1e9 + 7;

using namespace std;

int n, k;
int a[N];
int t1[N][20];
int t2[N][20];
long long s1[N][20];
long long s2[N][20];

int main()
{
        ios_base::sync_with_stdio(0);

        //freopen("input.txt", "r", stdin);
        //freopen("output.txt", "w", stdout);

        cin >> n >> k;
        for(int i = 1; i <= n; i++){
                cin >> a[i];
        }
        for(int i = 1, j = 0, h = 1; i <= n; i++){
                while(a[i] - a[j + 1] >= k){
                        j += 1;
                }
                while(h <= n && a[h] - a[i] < k){
                        h += 1;
                }
                t1[i][0] = s1[i][0] = j;
                t2[i][0] = s2[i][0] = h;
        }
        t2[n + 1][0] = n + 1;
        for(int i = 1; i < 20; i++){
                for(int j = 0; j <= n + 1; j++){
                        t1[j][i] = t1[t1[j][i - 1]][i - 1];
                        t2[j][i] = t2[t2[j][i - 1]][i - 1];
                        s1[j][i] = s1[j][i - 1] + s1[t1[j][i - 1]][i - 1];
                        s2[j][i] = s2[j][i - 1] + s2[t2[j][i - 1]][i - 1];
                }
        }
        int q;
        cin >> q;
        while(q--){
                int l, r;
                cin >> l >> r;
                int x = l, y = r;
                long long res = r - l + 1;
                for(int i = 19; i >= 0; i--){
                        if(t2[x][i] <= r){
                                assert(t1[y][i] >= l);
                                res -= s2[x][i];
                                res += s1[y][i];
                                res += (1 << i);
                                x = t2[x][i];
                                y = t1[y][i];
                        }
                }
                cout << res << "\n";
        }
}