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 <iostream>
#include <set>
#include <vector>

#define _overload(_1, _2, _3, name, ...) name
#define _rep(i, n)                       _range(i, 0, n)
#define _range(i, a, b)                  for (int i = int(a); i < int(b); ++i)
#define rep(...)                         _overload(__VA_ARGS__, _range, _rep, )(__VA_ARGS__)

#define _rrep(i, n)      _rrange(i, n, 0)
#define _rrange(i, a, b) for (int i = int(a) - 1; i >= int(b); --i)
#define rrep(...)        _overload(__VA_ARGS__, _rrange, _rrep, )(__VA_ARGS__)

using ll = long long;
using namespace std;

const int limit = 200010;
int nxt[18][limit];
ll nsum[18][limit];

int prv[18][limit];
ll psum[18][limit];


int main(void) {
    ll n, d;
    cin >> n >> d;
    vector<int> x(n);
    rep(i, n) cin >> x[i];

    {
        int pos = 0;
        rep(i, n) {
            while (pos < n and x[pos] - x[i] < d) {
                pos++;
            }
            nxt[0][i] = pos;
            nsum[0][i] += i;
        }
    }

    {
        int pos = n - 1;
        rrep(i, n) {
            while (pos >= 0 and x[i] - x[pos] < d) {
                pos--;
            }
            prv[0][i] = pos;
            psum[0][i] += i;
        }
    }

    rep(k, 1, 18) {
        rep(i, n) {
            nxt[k][i]  = (nxt[k - 1][i] == n) ? n : nxt[k - 1][nxt[k - 1][i]];
            nsum[k][i] = (nxt[k - 1][i] == n)
                             ? nsum[k - 1][i]
                             : nsum[k - 1][nxt[k - 1][i]] + nsum[k - 1][i];

            prv[k][i]  = (prv[k - 1][i] == -1) ? -1 : prv[k - 1][prv[k - 1][i]];
            psum[k][i] = (prv[k - 1][i] == -1)
                             ? psum[k - 1][i]
                             : psum[k - 1][prv[k - 1][i]] + psum[k - 1][i];
        }
    }

    int q;
    cin >> q;

    rep(loop, q) {
        int l, r;
        cin >> l >> r;
        l--, r--;

        ll num  = 1LL;
        int cur = l;
        rrep(k, 18) {
            if (nxt[k][cur] <= r) {
                cur = nxt[k][cur];
                num += 1 << k;
            }
        }

        ll ans = num;

        int cl = l;
        int cr = r;
        rrep(k, 18) {
            if (num >> k & 1) {
                ans -= nsum[k][cl];
                cl = nxt[k][cl];
                ans += psum[k][cr];
                cr = prv[k][cr];
            }
        }

        cout << ans << endl;
    }


    return 0;
}