#include <iostream>
#include <iomanip>
#include<math.h>
#include<list>
#include <algorithm>
#include<set>
#include<vector>
#include<math.h>
#include<map>
#include<string>
#include <numeric>
#include <queue>
#include <sstream>
#include <bitset>
#include<stack>
#include<functional>
#include<unordered_set>
#include<unordered_map>
#include<atcoder/all>

using namespace std;
using ll = long long;
using vll = vector<long long>;
using sll = set<long long>;
const long long ll_max = 9223372036854775807;
const long long ll_min = -9223372036854775807;
const int int_max = 2147483647;

template<typename T>
T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template<typename T>
T lcm(T a, T b) { return a / gcd(a, b) * b; }

template<typename T>
map<T, T> getPrimeFactor(T n) {
	map<T, T> ret;
	for (T i = 2; i * i <= n; ++i) {
		while (n % i == 0) {
			ret[i] ++;
			n /= i;
		}
	}
	if (n != 1) ret[n] = 1;
	return ret;
}

template<typename T>
bool IsPrimeNumber(T num)
{
	if (num <= 2) return true;
	else if (num % 2 == 0) return false;

	double sqrtNum = sqrt(num);
	for (int i = 3; i <= sqrtNum; i += 2)
	{
		if (num % i == 0)
		{
			return false;
		}
	}

	return true;
}

template<typename T>
T modinv(T a, T m) {
	T b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}



// 繰り返し二乗法
ll pow2(ll n, ll p, ll mod) {
	if (p == 0 || n == 1) {
		return 1;
	}
	else {
		ll ret = pow2(n * n % mod, p / 2, mod);
		if (p % 2 == 1) {
			ret *= n;
		}
		return ret % mod;
	}
}


#define rep(i,s,e) for(ll i=s;i<e;i++)
#define repeq(i,s,e) for(ll i=s;i<=e;i++)


ll n, k;
vll a;

map<ll, ll> dic;
ll func(ll idx) {
	auto itr = dic.find(idx);
	if (itr != dic.end()) {
		return itr->second;
	}
	ll ret = 1;
	rep(i, idx + 1, n) {
		if (abs(a[i] - a[idx]) <= k) {
			ret = max(ret, 1 + func(i));
		}
	}
	dic[idx] = ret;

	return ret;
}

int main() {
	cin >> n >> k;
	a.resize(n);
	rep(i, 0, n) {
		cin >> a[i];
	}

	ll ans = 0;
	for (ll i = n - 1; i >= 0; i--) {
		ans = max(ans, func(i));
	}

	cout << ans << endl;
	return 0;
}
