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: Wrong Answer)

#include <iostream>
#include <vector>
using namespace std;
struct segK{
    int n;
    vector<long long> dat;
    long long MAX;
    segK(int n_){
        n = 1;
        MAX = 0;
        while(n < n_) n *= 2;
        dat.resize(2 * n);
        for(int i = 0; i < 2 * n; i++) dat[i] = MAX;
    }
    void update(int k, int a){
        k += n;
        dat[k] = a;
        while(k > 0){
            k = k / 2;
            dat[k] = max(dat[k * 2], dat[k * 2 + 1]);
        }
    }
    long long query(int a, int b, int k = 0, int l = 0, int r = -1){
        if(r < 0) r = n;
        if(r <= a || b <= l) return MAX;
        if(a <= l && r <= b) return dat[k];
        else{
            long long vl = query(a, b, k * 2, l, (l + r) / 2);
            long long vr = query(a, b, k * 2 + 1, (l + r) / 2, r);
            return max(vl, vr);
        }
    }
};
int main(){
    int N, K;
    cin >> N >> K;
    vector<int> A(N);
    for(int i = 0; i < N; i++) cin >> A[i];
    segK seg(300100);
    for(int i = 0; i < N; i++){
        if(i == 0) seg.update(A[i], 1);
        else{
            int l = max(0, A[i] - K);
            int r = min(300050, A[i] + K);
            int res = seg.query(l, r + 1);
            seg.update(A[i], res + 1);
        }
    }
    cout << seg.query(0, 300100) << endl;
}