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: Wrong Answer)

#include <bits/stdc++.h>

#define pb push_back
#define PQ priority_queue
#define make_unique(x) sort((x).begin(), (x).end()); (x).erase(unique((x).begin(), (x).end()), (x).end())

typedef long long ll;
typedef long double ld;

//__builtin_popcount(x)

using namespace std;

/*const int mod = 1e9 + 7;

ll add(ll a, ll b) {
    ll res = a + b;
    if (res >= mod) res -= mod;
    return res;
}

ll sub(ll a, ll b) {
    ll res = a - b + mod;
    if (res >= mod) res -= mod;
    return res;
}

ll mul(ll a, ll b) {
    return (((a % mod) * (b % mod)) % mod);
}

int powm(int a, long long n) {
	if (n == 0) {
		return 1;
	}
	else if (n & 1) {
		return mul(a, powm(a, n - 1));
	}
	else {
		return powm(mul(a, a), n / 2);
	}
}*/

/* Z-FUNCTION
ll l = 0, r = 0, n = (int)s.size();
vector<int> z(n, 0);
for (int i = 1; i < n; i++) {
    if (r >= i) {
        z[i] = min(1ll*z[i-l], r-i+1);
    }
    while (z[i] + i < n && s[z[i]] == s[z[i] + i]) z[i]++;
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
}
*/

/*vector<ll> find_dividers(ll x) {
    vector<ll> dividers;

    for (ll i = 1; i <= sqrt(x); i++) {
        if (x % i == 0) {
           dividers.push_back(i);
            if (1ll*i * i != x) {
                dividers.push_back(1ll*x / 1ll*i);
            }
        }
    }

    return dividers;
}*/

/*int nok(int a, int b) {
    return a / __gcd(a, b) * b;
}*/

// ceil(a/b) = (a+b-1)/b

void solve() {
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<int> ab;
    int total = 0;
    for (int i = 0; i < n-1; i++) {
        if (abs(a[i] - a[i + 1]) <= k) total++;
    }
    cout << total+1 << endl;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    //cout.setf(ios::fixed);
    //cout.precision(10);

    int t=1;
    //cin >> t;
    while (t--) {
        solve();
    }

    return 0;
}