Problem:
Score : 400 points

Problem Statement

You are given a sequence A_1, A_2, ..., A_N and an integer K.

Print the maximum possible length of a sequence B that satisfies the following conditions:

B is a (not necessarily continuous) subsequence of A.

For each pair of adjacents elements of B, the absolute difference of the elements is at most K.

Constraints

1 \leq N \leq 300,000

0 \leq A_i \leq 300,000

0 \leq K \leq 300,000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1
A_2
:
A_N

Output

Print the answer.

Sample Input 1

10 3
1
5
4
3
8
6
9
7
2
4

Sample Output 1

7

For example, B = (1, 4, 3, 6, 9, 7, 4) satisfies the conditions.

It is a subsequence of A = (1, 5, 4, 3, 8, 6, 9, 7, 2, 4).

All of the absolute differences between two adjacent elements (|1-4|, |4-3|, |3-6|, |6-9|, |9-7|, |7-4|) are at most K = 3.

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
#include <stack>
#include <functional>

// AtCoder
// #include <atcoder/all>
// using namespace atcoder;

#ifdef LOCAL
    #define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
    #define eprintf(...)
#endif

#define rep_(i, a_, b_, a, b, ...) for (int i = (a), i##_len = (b); i < i##_len; ++i)
#define rep(i, ...) rep_(i, __VA_ARGS__, __VA_ARGS__, 0, __VA_ARGS__)
#define reprev_(i, a_, b_, a, b, ...) for (int i = (b-1), i##_min = (a); i >= i##_min; --i)
#define reprev(i, ...) reprev_(i, __VA_ARGS__, __VA_ARGS__, 0, __VA_ARGS__)
#define all(x) (x).begin(), (x).end()
template <class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template <class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return 1; } return 0; }
#define fls(x) (64 - __builtin_clzll(x))
#define pcnt(x) __builtin_popcountll(x)
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair <int,int> P;
typedef long double ld;

template< typename Monoid >
struct SegmentTree {
    using F = function< Monoid(Monoid, Monoid) >;
    int sz;
    vector< Monoid > seg;
    const F f;
    const Monoid M1;
    SegmentTree(int n, const F f, const Monoid &M1) : f(f), M1(M1) {
        sz = 1;
        while(sz < n) sz <<= 1;
        seg.assign(2 * sz, M1);
    }
    void set(int k, const Monoid &x) {
        seg[k + sz] = x;
    }
    void build() {
        for(int k = sz - 1; k > 0; k--) {
            seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
        }
    }
    void update(int k, const Monoid &x) {
        k += sz;
        seg[k] = x;
        while(k >>= 1) {
            seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
        }
    }
    Monoid query(int a, int b) {
        Monoid L = M1, R = M1;
        for(a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
            if(a & 1) L = f(L, seg[a++]);
            if(b & 1) R = f(seg[--b], R);
        }
        return f(L, R);
    }
    Monoid operator[](const int &k) const {
        return seg[k + sz];
    }
    template< typename C >
    int find_subtree(int a, const C &check, Monoid &M, bool type) {
        while(a < sz) {
            Monoid nxt = type ? f(seg[2 * a + type], M) : f(M, seg[2 * a + type]);
            if(check(nxt)) a = 2 * a + type;
            else M = nxt, a = 2 * a + 1 - type;
        }
        return a - sz;
    }
    template< typename C >
    int find_first(int a, const C &check) {
        Monoid L = M1;
        if(a <= 0) {
            if(check(f(L, seg[1]))) return find_subtree(1, check, L, false);
            return -1;
        }
        int b = sz;
        for(a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
            if(a & 1) {
                Monoid nxt = f(L, seg[a]);
                if(check(nxt)) return find_subtree(a, check, L, false);
                L = nxt;
                ++a;
            }
        }
        return -1;
    }
    template< typename C >
    int find_last(int b, const C &check) {
        Monoid R = M1;
        if(b >= sz) {
            if(check(f(seg[1], R))) return find_subtree(1, check, R, true);
            return -1;
        }
        int a = sz;
        for(b += sz; a < b; a >>= 1, b >>= 1) {
            if(b & 1) {
                Monoid nxt = f(seg[--b], R);
                if(check(nxt)) return find_subtree(b, check, R, true);
                R = nxt;
            }
        }
        return -1;
    }
};

int main (void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, k; cin >> n >> k;
    vector<int> a(n);
    rep (i, n) cin >> a[i];
    // SegmentTree<ll> segmin(n, [](ll a, ll b){return min(a, b);}, 10000000);
    // SegmentTree<ll> segmax(n, [](ll a, ll b){return max(a, b);}, 0);
    SegmentTree<int> seg(300010, [](int a, int b) { return min(a, b);}, 1000000);
    vector<P> p;
    vector<ll> dp(n + 1, 1);
    reprev (i, n) {
        if (int x = seg.query(max(0, a[i] - k), a[k] + 1); x < n) {
            p.emplace_back(i, x);
        }
        if (int x = seg.query(a[k], min(300010, a[i] + k + 1)); x < n) {
            p.emplace_back(i, x);
        }
        seg.update(a[i], i);
    }
    reverse(all(p));
    for (auto [a, b] : p) {
        eprintf("%d %d\n", a, b);
        chmax(dp[b], dp[a] + 1);
    }
    ll ans = 0;
    rep (i, n) chmax(ans, dp[i]);
    cout << ans << "\n";

    /***************************************************/
    /* Use "submit code (with ACL v1)" when using ACL! */
    /***************************************************/
    return 0;
}