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;
#ifdef tabr
#include "library/debug.cpp"
#else
#define debug(...)
#endif

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    vector<int> x(n);
    for (int i = 0; i < n; i++) {
        cin >> x[i];
    }
    vector<vector<int>> ld(20, vector<int>(n));
    vector<vector<int>> rd(20, vector<int>(n));
    vector<vector<long long>> ls(20, vector<long long>(n));
    vector<vector<long long>> rs(20, vector<long long>(n));
    for (int i = 0; i < n; i++) {
        ld[0][i] = upper_bound(x.begin(), x.end(), x[i] - k) - x.begin() - 1;
        rd[0][i] = lower_bound(x.begin(), x.end(), x[i] + k) - x.begin();
        ls[0][i] = ld[0][i];
        rs[0][i] = rd[0][i];
    }
    for (int i = 1; i < 20; i++) {
        for (int j = 0; j < n; j++) {
            ld[i][j] = (ld[i - 1][j] == -1 ? -1 : ld[i - 1][ld[i - 1][j]]);
            rd[i][j] = (rd[i - 1][j] == n ? n : rd[i - 1][rd[i - 1][j]]);
            ls[i][j] = (ld[i - 1][j] == -1 ? 0 : ls[i - 1][j] + ls[i - 1][ld[i - 1][j]]);
            rs[i][j] = (rd[i - 1][j] == n ? 0 : rs[i - 1][j] + rs[i - 1][rd[i - 1][j]]);
        }
    }
    int q;
    cin >> q;
    while (q--) {
        int l, r;
        cin >> l >> r;
        if (l == r) {
            cout << 1 << '\n';
            continue;
        }
        l--, r--;
        long long ans = r - l + 1;
        int pl = l, pr = r;
        for (int i = 19; i >= 0; i--) {
            if (rd[i][pl] <= r) {
                ans -= rs[i][pl];
                pl = rd[i][pl];
                ans += 1 << i;
            }
            if (ld[i][pr] >= l) {
                ans += ls[i][pr];
                pr = ld[i][pr];
            }
        }
        cout << ans << '\n';
    }
    return 0;
}