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;

const int max_n = 200222, inf = 1000111222;
const int max_lev = 18;

int n, k, q, a[max_n];
int nxt[2][max_n][max_lev];
long long sum[2][max_n][max_lev];

void calc_nxt(int tp) {
    nxt[tp][n][0] = n;
    for (int i = n - 1; i >= 0; --i) {
        nxt[tp][i][0] = nxt[tp][i + 1][0];
        sum[tp][i][0] = i;
        if (tp) {
            sum[tp][i][0] = n - i;
        }
        while (a[nxt[tp][i][0] - 1] - a[i] >= k) {
            --nxt[tp][i][0];
        }
    }
    for (int lev = 1; lev < max_lev; ++lev) {
        for (int i = 0; i <= n; ++i) {
            nxt[tp][i][lev] = nxt[tp][nxt[tp][i][lev - 1]][lev - 1];
            sum[tp][i][lev] = sum[tp][i][lev - 1] + sum[tp][nxt[tp][i][lev - 1]][lev - 1];
        }
    }
}

void rev() {
    reverse(a, a + n);
    for (int i = 0; i < n; ++i) {
        a[i] *= -1;
    }
}

long long get_sum(int tp, int pos, int r) {
    long long res = 0;
    for (int i = max_lev - 1; i >= 0; --i) {
        if (nxt[tp][pos][i] <= r) {
            res += sum[tp][pos][i];
            pos = nxt[tp][pos][i];
        }
    }
    res += sum[tp][pos][0];
    return res;
}

int main() {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; ++i) {
        scanf("%d", &a[i]);
    }
    calc_nxt(0);
    rev();
    calc_nxt(1);
    rev();
    scanf("%d", &q);
    while (q--) {
        int l, r;
        scanf("%d%d", &l, &r);
        --l;
        --r;
        long long s1 = get_sum(0, l, r);
        long long s2 = get_sum(1, n - r - 1, n - l - 1);
        //cout << s1 << " " << s2 << ": ";
        int ans = s2 - s1;
        printf("%d\n", ans);
    }
    return 0;
}