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 with AC Library v1.1) (judge verdict: Accepted)

#pragma region header
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include<atcoder/all>
using namespace std;
using namespace atcoder;
#define int long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rev(i, n) for(int i = (int)(n - 1); i >= 0; i--)
#define rev1(i, n) for(int i = (int)(n); i > 0; i--)
#define pb push_back
#define all(v) (v).begin(), (v).end()
#define resort(v) sort((v).rbegin(), (v).rend())
#define vi vector<int>
#define vvi vector<vector<int>>
#define vc vector<char>
#define vvc vector<vector<char>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
using ll = long long;
using P = pair<int, int>;
/* ----------------よく使う数字や配列----------------- */
int dx[] = {1,0,-1,0};
int dy[] = {0,1,0,-1};
constexpr ll mod = 1e9+7;
constexpr int inf = INT32_MAX/2;
constexpr ll INF = LLONG_MAX/2;
constexpr long double eps = DBL_EPSILON;
constexpr long double pi = 3.141592653589793238462643383279;
/* ----------------------end----------------------- */

/* --------------------テンプレート------------------ */
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
/* ----------------------end----------------------- */

/* --------------------ライブラリ-------------------- */
ll fact(int i) {       //階乗
    if (i == 0) return 1;
    return (fact(i - 1)) * i % mod;
}
ll gcd(ll a, ll b) {        //最大公約数
    if(b == 0) return a;
    return gcd(b, a % b); 
}
ll lcm(ll a, ll b) {      //最小公倍数
    return a / gcd(a, b) * b ;
}
int keta(ll n) {        //桁数を求める
    if(n == 0) return 1;
    int count = 0;
    while(n != 0) {
        n /= 10;
        count++;
    }
    return count;
}
ll ketasum(ll n) {    //各桁の和
    ll sum = 0;
    while(n != 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}
bool isprime(int n) { //素数判定
    if(n==1)return false;
    if(n==2)return true;
    for(int i = 2;i<=sqrt(n);i++) {
        if(n%i==0) return false;
    }
    return true;
}
/* ----------------------end----------------------- */
#pragma endregion
int op(int a,int b) {return max(a,b);}
int e() {return 0;}
signed main() {
    int n,k;cin >> n >> k;
    vi v(n);
    rep(i, n) cin >> v[i];
    segtree<int,op,e> dp(300300);
    rep(i, n) {
        int x = v[i];
        int l = max(0LL,x-k),r=min(300300LL,x+k+1);
        dp.set(x,dp.prod(l,r)+1);
    }
    cout << dp.prod(0,300300) << endl;
    return 0;
}