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;

typedef long long ll;
const int N = (int) 2e5 + 7;
const int K = 18;
int n;
int dif;
int q;
int a[N];
int u[K][N];
ll f[K][N];
int u2[K][N];
ll f2[K][N];

ll compute1(int x, int lim)
{
    ll sol = x - 1;
    for (int k = K - 1; k >= 0; k--)
    {
        if (u[k][x] <= lim)
        {
            sol += f[k][x] - (1 << k);
            x = u[k][x];
        }
    }
    return sol;
}

ll compute2(int x, int lim)
{
    ll sol = x;
    for (int k = K - 1; k >= 0; k--)
    {
        if (lim <= u2[k][x])
        {
            sol += f2[k][x];
            x = u2[k][x];
        }
    }
    return sol;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);


    cin >> n >> dif;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    a[n + 1] = a[n] + dif;
    int j = 1;
    for (int i = 1; i <= n; i++)
    {
        while (a[j] - a[i] < dif)
        {
            j++;
        }
        u[0][i] = j;
        f[0][i] = j;
    }
    a[0] = a[1] - dif;
    j = n;
    for (int i = n; i >= 1; i--)
    {
        while (a[i] - a[j] < dif)
        {
            j--;
        }
        u2[0][i] = j;
        f2[0][i] = j;
    }
    for (int k = 1; k < K; k++)
    {
        for (int i = 1; i <= n; i++)
        {
            if (u[k - 1][i] == n + 1)
            {
                u[k][i] = n + 1;
            }
            else
            {
                f[k][i] = f[k - 1][i] + f[k - 1][u[k - 1][i]];
                u[k][i] = u[k - 1][u[k - 1][i]];
            }
            /// boooo
            if (u2[k - 1][i] == 0)
            {
                u2[k][i] = 0;
            }
            else
            {
                f2[k][i] = f2[k - 1][i] + f2[k - 1][u2[k - 1][i]];
                u2[k][i] = u2[k - 1][u2[k - 1][i]];
            }
        }
    }
    cin >> q;
    while (q--)
    {
        int l, r, best = 0, sol = 0;
        cin >> l >> r;
        cout << compute2(r, l) - compute1(l, r) << "\n";
    }
    return 0;

}