#include <bits/stdc++.h>
using namespace std;
struct Segment { // max 0-segment
bool allZero = false;
int lft = 0;
int rgt = 0;
int opt = 0;
Segment(int c = 0) {
if (c == 0) {
allZero = 1;
lft = 1;
rgt = 1;
opt = 1;
} else {
allZero = 0;
lft = 0;
rgt = 0;
opt = 0;
}
}
friend Segment operator+(const Segment &a, const Segment &b) {
Segment c(0);
c.allZero = a.allZero && b.allZero;
c.lft = a.lft + (a.allZero ? b.lft : 0);
c.rgt = b.rgt + (b.allZero ? a.rgt : 0);
c.opt = max({a.opt, b.opt, a.rgt + b.lft});
return c;
}
};
class MaxSubsegment {
public:
int sz;
vector<Segment> tree;
MaxSubsegment(int sz) : sz(sz) {
tree.resize(2 * sz);
for (int i = sz - 1; i >= 0; i--) {
tree[i] = tree[i * 2] + tree[i * 2 + 1];
}
}
void Modify(int p, int x) {
tree[p += sz] = Segment(x);
for (p /= 2; p > 0; p /= 2) {
tree[p] = tree[p * 2] + tree[p * 2 + 1];
}
}
Segment Query(int l, int r) {
Segment lft(1), rgt(1);
for (l += sz, r += sz + 1; l < r; l /= 2, r /= 2) {
if (l & 1) lft = lft + tree[l++];
if (r & 1) rgt = tree[--r] + rgt;
}
return lft + rgt;
}
};
class SegTree {
public:
int sz;
vector<int> tree;
SegTree(int sz) : sz(sz), tree(2 * sz, -1e9) {}
int Query(int l, int r) {
int z = 0;
for (l += sz, r += sz + 1; l < r; l /= 2, r /= 2) {
if (l & 1) z = max(z, tree[l++]);
if (r & 1) z = max(z, tree[--r]);
}
return z;
}
void Modify(int p, int x) {
tree[p += sz] = x;
for (p /= 2; p > 0; p /= 2) {
tree[p] = max(tree[p * 2], tree[p * 2 + 1]);
}
}
};
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int N, D;
cin >> N >> D;
vector<int> A(N);
for (int i = 0; i < N; i++) {
cin >> A[i];
}
{ // Coordinate Compression
vector<int> coords = A;
sort(begin(coords), end(coords));
coords.resize(unique(begin(coords), end(coords)) - begin(coords));
for (auto &a : A) {
a = lower_bound(begin(coords), end(coords), a) - begin(coords);
}
}
// dp[pos][max_element]
// find maximum value on dp[pos-D...pos][0...A[pos] - 1] -> dp[pos][A[pos]] = 1 + best
// for x >= A[pos]: dp[pos-D...pos][x] -> dp[pos][x]
//
// If we group all occurrences of value val which has distance <= D, then
// dp[i][val] <= dp[j][val] for i < j (since we can retake "val").
// We can transform the sequence such that this condition is fullfilled without
// the distance <= D condition.
//
// How? We scan from least valued element, then when we arrive at "val", group them
// which has distance <= D. Afterwards, we reassign values s.t. the ones in the back
// has lower value than the ones in the front.
//
// After this transformation, dp[i][val] <= dp[j][val] for i < j. So, we only need
// to store last occurrence of dp[i][val]. This yields O(N^2).
//
// For a value "val", since all adjacent occurrence has distance <= D, then the value
// "val" is valid only for some subsegment of the array. We can precompute this, then
// we only need to consider "active" values when transitioning. Afterwards, we don't
// need the array last[], and it's basically a simple DP optimized with segment tree.
//
// Time complexity: O(N log N).
{ // Transformation s.t. dp[i][val] <= dp[j][val] for i < j
int value = 0;
vector<vector<int>> occ(N);
for (int i = 0; i < N; i++) {
occ[A[i]].emplace_back(i);
}
for (int v = 0; v < N; v++) {
int cnt = 0;
for (int i = 0; i < int(occ[v].size()); i++) {
int j = i;
while (j + 1 < int(occ[v].size()) && occ[v][j + 1] - occ[v][j] <= D) {
j += 1;
}
cnt += 1;
i = j;
}
value += cnt;
cnt = 0;
for (int i = 0; i < int(occ[v].size()); i++) {
int j = i;
while (j + 1 < int(occ[v].size()) && occ[v][j + 1] - occ[v][j] <= D) {
j += 1;
}
cnt += 1;
for (int k = i; k <= j; k++) {
A[occ[v][k]] = value - cnt;
}
i = j;
}
}
assert(*max_element(begin(A), end(A)) < N);
}
const auto DP_N2_Old = [&](int N, int D, vector<int> A) {
vector<int> dp(N, -1e9);
vector<int> last(N, -1e9);
for (int pos = 0; pos < N; pos++) {
for (int i = A[pos]; i < N; i++) {
if (pos - last[i] <= D) {
last[i] = pos;
}
}
last[A[pos]] = pos;
dp[A[pos]] = max(dp[A[pos]], 1);
for (int i = 0; i < A[pos]; i++) {
if (pos - last[i] <= D) {
dp[A[pos]] = max(dp[A[pos]], dp[i] + 1);
}
}
}
int ans = 0;
for (int i = A[N - 1]; i < N; i++) {
if (N - 1 - last[i] <= D) {
ans = max(ans, dp[i]);
}
}
return ans;
};
const auto Solve = [&](int N, int D, vector<int> A) {
vector<int> startValid(N, -1);
vector<int> endValid(N, -1);
for (int i = 0; i < N; i++) {
if (startValid[A[i]] == -1) {
startValid[A[i]] = i;
}
endValid[A[i]] = i;
}
vector<vector<int>> occ(N);
for (int i = 0; i < N; i++) {
occ[A[i]].emplace_back(i);
}
MaxSubsegment maxSeg(N);
for (int i = 0; i < N; i++) if (endValid[i] != -1) {
for (auto j : occ[i]) {
maxSeg.Modify(j, 1);
}
int lo = endValid[i];
int hi = N - 1;
while (lo < hi) {
int md = (lo + hi) / 2;
if (maxSeg.Query(endValid[i], md).opt >= D) {
hi = md;
} else {
lo = md + 1;
}
}
endValid[i] = lo;
}
SegTree dpSeg(N);
vector<vector<int>> endEvent(N + 1);
for (int i = 0; i < N; i++) {
endEvent[endValid[i] + 1].emplace_back(i);
}
for (int pos = 0; pos < N; pos++) {
for (auto j : endEvent[pos]) {
dpSeg.Modify(j, -1e9);
}
int val = max({dpSeg.Query(A[pos], A[pos]), 1, 1 + dpSeg.Query(0, A[pos] - 1)});
dpSeg.Modify(A[pos], val);
}
return dpSeg.Query(0, N - 1);
};
cout << Solve(N, D, A) << '\n';
return 0;
}
