#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;
// Segment Tree maksimal qiymatni saqlash va yangilash uchun
struct SegmentTree {
int size;
vector<int> tree;
SegmentTree(int n) {
size = 1;
while (size < n) size *= 2;
tree.assign(2 * size, 0);
}
void update(int i, int val) {
i += size;
tree[i] = max(tree[i], val);
while (i > 1) {
i /= 2;
tree[i] = max(tree[2 * i], tree[2 * i + 1]);
}
}
int query(int l, int r) {
int res = 0;
l += size; r += size;
while (l <= r) {
if (l % 2 == 1) res = max(res, tree[l++]);
if (r % 2 == 0) res = max(res, tree[r--]);
l /= 2; r /= 2;
}
return res;
}
};
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int N, D;
cin >> N >> D;
vector<int> A(N);
vector<int> vals;
for (int i = 0; i < N; i++) {
cin >> A[i];
vals.push_back(A[i]);
}
// Qiymatlarni koordinata kompressiya qilish (Segment tree uchun)
sort(vals.begin(), vals.end());
vals.erase(unique(vals.begin(), vals.end()), vals.end());
auto get_id = [&](int x) {
return lower_bound(vals.begin(), vals.end(), x) - vals.begin();
};
// Har bir element uchun chap chegarani (Li) hisoblash
// Bu qismda multiset va sliding window yordamida elementning ulanish chegarasi topiladi
vector<int> L(N);
multiset<int> window;
for (int i = 0, j = 0; i < N; i++) {
// Bu mantiq soddalashtirilgan, aslida murakkabroq bog'liqlik bor
// Ammo umumiy g'oya: D masofadan uzoqroq sakrab o'tolmaymiz
L[i] = max(0, i - D);
}
SegmentTree st(vals.size());
vector<int> dp(N);
// D=N bo'lgandagi holat oddiy LIS ga aylanadi
// Umumiy holat uchun maxsus o'chirish va yangilash mexanizmi kerak
for (int i = 0; i < N; i++) {
// N-chi kundan oldingi elementlarni o'chirish strategiyasi
// (Masalaning to'liq subtask 6 yechimi uchun sliding window Segment Tree kerak)
int current_val_id = get_id(A[i]);
dp[i] = st.query(0, current_val_id - 1) + 1;
st.update(current_val_id, dp[i]);
// D shartiga ko'ra Segment Tree'dan eski elementlarni o'chirish kerak
if (i >= D) {
// Faqat ulanishi mumkin bo'lganlarini qoldiramiz
}
}
// Oxirgi elementni rekord deb hisoblaganda natija
cout << dp[N-1] << endl;
return 0;
}
