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;
using lint = long long;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template <typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template <typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template <typename V, typename T> void ndfill(V &x, const T &val) { x = val; }
template <typename V, typename T> void ndfill(vector<V> &vec, const T &val) { for (auto &v : vec) ndfill(v, val); }
template <typename T> bool chmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template <typename T> bool chmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template <typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template <typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
template <typename T> vector<T> srtunq(vector<T> vec) { sort(vec.begin(), vec.end()), vec.erase(unique(vec.begin(), vec.end()), vec.end()); return vec; }
template <typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto &v : vec) is >> v; return is; }
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { os << '['; for (auto v : vec) os << v << ','; os << ']'; return os; }
#if __cplusplus >= 201703L
template <typename... T> istream &operator>>(istream &is, tuple<T...> &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); return is; }
template <typename... T> ostream &operator<<(ostream &os, const tuple<T...> &tpl) { std::apply([&os](auto &&... args) { ((os << args << ','), ...);}, tpl); return os; }
#endif
template <typename T> ostream &operator<<(ostream &os, const deque<T> &vec) { os << "deq["; for (auto v : vec) os << v << ','; os << ']'; return os; }
template <typename T> ostream &operator<<(ostream &os, const set<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T> ostream &operator<<(ostream &os, const multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa) { os << '(' << pa.first << ',' << pa.second << ')'; return os; }
template <typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
template <typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
#ifdef HITONANODE_LOCAL
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl
#else
#define dbg(x)
#endif

int main()
{
    int N, K;
    cin >> N >> K;
    vector<int> X(N);
    cin >> X;
    vector<int> prv(N), nxt(N);
    REP(i, N) nxt[i] = lower_bound(ALL(X), X[i] + K) - X.begin();
    REP(i, N) prv[i] = upper_bound(ALL(X), X[i] - K) - X.begin() - 1;

    REP(i, N) if (prv[i] < 0) prv[i] = N;

    constexpr int D = 20;
    vector<vector<int>> dnxt(D, vector<int>(N + 1, N));
    vector<vector<lint>> sumnxt(D, vector<lint>(N + 1));
    vector<vector<int>> dprv(D, vector<int>(N + 1, N));
    vector<vector<lint>> sumprv(D, vector<lint>(N + 1));
    dnxt[0] = nxt;
    dprv[0] = prv;
    REP(i, N + 1)
    {
        sumnxt[0][i] = N - i; // dnxt[0][i];
        sumprv[0][i] = N - i; // dprv[0][i];
    }
    REP(d, D - 1)
    {
        REP(i, N)
        {
            dnxt[d + 1][i] = dnxt[d][dnxt[d][i]], sumnxt[d + 1][i] = sumnxt[d][i] + sumnxt[d][dnxt[d][i]];
            dprv[d + 1][i] = dprv[d][dprv[d][i]], sumprv[d + 1][i] = sumprv[d][i] + sumprv[d][dprv[d][i]];
        }
    }

    dbg(dnxt[0]);
    dbg(dnxt[1]);
    dbg(dnxt[2]);
    dbg(sumnxt[0]);
    dbg(sumnxt[1]);
    dbg(sumnxt[2]);
    dbg(sumprv[0]);
    dbg(sumprv[1]);
    dbg(sumprv[2]);
    int Q;
    cin >> Q;
    while (Q--)
    {
        int l, r;
        cin >> l >> r;
        l--, r--;
        int now = l, nowinv = r;
        int step = 0;

        IREP(d, D)
        {
            if (dnxt[d][now] <= r)
            {
                step += 1 << d;
                now = dnxt[d][now];
            }
        }
        step++;

        now = l, nowinv = r;
        lint ret1 = 0, ret2 = 0;
        IREP(d, D) if ((step >> d) & 1)
        {
                ret1 += sumnxt[d][now];
                ret2 += sumprv[d][nowinv];
                now = dnxt[d][now];
                nowinv = dprv[d][nowinv];
        }
        cout << ret1 - ret2 + step << '\n';
        dbg(step);
        dbg(ret1);
        dbg(ret2);
    }
}