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;

#ifdef LOCAL
#define dbg(...) debug(#__VA_ARGS__, __VA_ARGS__);
#else
#define dbg(...) 17;
#endif

template<typename T, typename S> ostream& operator << (ostream &os, const pair<T, S> &p) { return os << "(" << p.first << ", " << p.second << ")"; }
template<typename C, typename T = decay<decltype(*begin(declval<C>()))>, typename enable_if<!is_same<C, string>::value>::type* = nullptr>
ostream& operator << (ostream &os, const C &c) { bool f = true; os << "{"; for (const auto &x : c) { if (!f) os << ", "; f = false; os << x; } return os << "}"; }
template<typename T> void debug(string s, T x) { cerr << s << " = " << x << "\n"; }
template<typename T, typename... Args> void debug(string s, T x, Args... args) { cerr << s.substr(0, s.find(',')) << " = " << x << " | "; debug(s.substr(s.find(',') + 2), args...); }

int main() {
    const int INF = 1e9;
    ios_base::sync_with_stdio(0); cin.tie(0);
    int n, k;
    cin >> n >> k;
    vector<int> x(n);
    set<pair<int, int>> pos;
    for (int i = 0; i < n; i++) {
        cin >> x[i];
        pos.emplace(x[i], i);
    }
    int d = 0;
    while ((1 << d) < n) {
        d++;
    }
    vector<vector<int>> fgo(n, vector<int>(d + 1, INF));
    vector<vector<ll>> fsum(n, vector<ll>(d + 1, -1));
    vector<vector<int>> bgo(n, vector<int>(d + 1, -INF));
    vector<vector<ll>> bsum(n, vector<ll>(d + 1, -1));
    for (int i = 0; i < n; i++) {
        auto it = pos.lower_bound({x[i] + k, -INF});
        if (it != pos.end()) {
            fgo[i][0] = (*it).second;
            fsum[i][0] = (*it).second;
        }
        it = pos.lower_bound({x[i] - k, INF});
        if (it != pos.begin()) {
            it = prev(it);
            bgo[i][0] = (*it).second;
            bsum[i][0] = (*it).second;
        }
    }
    for (int i = 0; i <= d; i++) {
        for(int j = 0; j < n; j++) {
            if (!(fgo[j][i] == INF || fgo[fgo[j][i]][i] == INF)) {
                fgo[j][i + 1] = fgo[fgo[j][i]][i];
                fsum[j][i + 1] = fsum[j][i] + fsum[fgo[j][i]][i];
            }
            if (!(bgo[j][i] == -INF || bgo[bgo[j][i]][i] == -INF)) {
                bgo[j][i + 1] = bgo[bgo[j][i]][i];
                bsum[j][i + 1] = bsum[j][i] + bsum[bgo[j][i]][i];
            }
        }
    }
    int q;
    cin >> q;
    while (q--) {
        int l, r;
        cin >> l >> r;
        l--, r--;
        int fcur = l;
        int bcur = r;
        ll ftot = l;
        ll btot = r;
        int tot = 1;
        for (int i = d; i >= 0; i--) {
            if (fgo[fcur][i] <= r) {
                ftot += fsum[fcur][i];
                tot += (1 << i);
                fcur = fgo[fcur][i];
            }
            if (bgo[bcur][i] >= l) {
                btot += bsum[bcur][i];
                bcur = bgo[bcur][i];
            }
        }
        cout << btot - ftot + tot << '\n';
    }
    return 0;
}