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, a, b) for (int i = a; i <= b; i++)
#define per(i, a, b) for (int i = a; i >= b; i--)
using namespace std;

typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef long long ll;

template <typename _T>
inline void read(_T &f) {
    f = 0; _T fu = 1; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') { fu = -1; } c = getchar(); }
    while (c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
    f *= fu;
}

template <typename T>
void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x < 10) putchar(x + 48);
    else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
    print(x); putchar(t);
}

const int N = 2e5 + 5;

ll suml[N][20], sumr[N][20];
int a[N], l[N][20], r[N][20];
int n, k, q;

int main() {
    read(n); read(k);
    for (int i = 1; i <= n; i++) read(a[i]);
    for (int i = 1; i <= n; i++) {
        l[i][0] = upper_bound(a + 1, a + n + 1, a[i] - k) - a - 1;
        r[i][0] = lower_bound(a + 1, a + n + 1, a[i] + k) - a;
        suml[i][0] = i; sumr[i][0] = i - 1;
    }
    r[n + 1][0] = n + 1;
    for (int j = 1; j <= 19; j++) {
        for (int i = 0; i <= n + 1; i++) {
            l[i][j] = l[l[i][j - 1]][j - 1];
            suml[i][j] = suml[i][j - 1] + suml[l[i][j - 1]][j - 1];
            r[i][j] = r[r[i][j - 1]][j - 1];
            sumr[i][j] = sumr[i][j - 1] + sumr[r[i][j - 1]][j - 1];
        }
    }
    read(q);
    while (q--) {
        int L, R; read(L); read(R);
        ll sumL = 0, sumR = 0;
        int nowL = L, nowR = R;
        for (int j = 19; j >= 0; j--) {
            if (r[nowL][j] <= R) {
                sumL += sumr[nowL][j];
                nowL = r[nowL][j];
            }
            if (l[nowR][j] >= L) {
                sumR += suml[nowR][j];
                nowR = l[nowR][j];
            }
        }
        print(sumR - sumL + nowR - nowL + 1, '\n');
    }
    return 0;
}