#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define MAXN (1000005)
ll N,D;
struct node1 {
ll s,e,m,v;
node1 *l, *r;
node1(ll _s,ll _e){
s = _s;
e = _e;
m = (s + e) / 2;
v = 0;
if(s != e){
l = new node1(s,m);
r = new node1(m + 1,e);
}
}
void update(ll x,ll nval){
if(s == e){
v = nval;
return;
}else{
if(x > m) r -> update(x,nval);
else l -> update(x,nval);
v = max(l -> v,r -> v);
}
}
ll bs(ll x){ //binary serach for leftmost position with v > x
if(s == e){
if(v <= x) return N + 5;
else return s;
}else{
if(l->v > x) return l -> bs(x);
else if(r->v > x) return r -> bs(x);
else return N + 5;
}
}
} *root1;
struct node2 {
ll s,e,m,lazy,v;
node2 *l, *r;
node2(ll _s,ll _e){
s = _s;
e = _e;
m = (s + e) / 2;
lazy = -1;
v = 0;
if(s != e){
l = new node2(s,m);
r = new node2(m + 1,e);
}
}
void propo(){
if(lazy == -1) return;
v = max(v,lazy);
if(s != e){
l->lazy = max(l->lazy,lazy);
r->lazy = max(r->lazy,lazy);
}
lazy = -1;
}
void update(ll x,ll y,ll nval){
propo();
if(s == x && e == y){
lazy = max(lazy,nval);
return;
}else{
if(x > m) r -> update(x,y,nval);
else if(y <= m) l -> update(x,y,nval);
else l -> update(x,m,nval), r -> update(m + 1,y,nval);
l -> propo(), r -> propo();
v = max(l -> v,r -> v);
}
}
ll query(ll x){
propo();
if(s == e){
return v;
}else{
if(x > m) return r -> query(x);
else return l -> query(x);
}
}
} *root2;
int main() {
ios_base::sync_with_stdio(false);cin.tie(0);
cin>>N>>D;
ll mini[N], rightmost[N];
pair<ll,ll> arr[N];
root1 = new node1(0,N + 5);
root2 = new node2(0,N + 5);
for(ll i = 0;i < N;i++){
cin>>arr[i].first;
arr[i].second = -i;
}
multiset<ll> s;
for(ll i = 0;i < N;i++){
s.insert(arr[i].first);
if(i >= D) s.erase(s.find(arr[i - D].first));
mini[i] = *s.begin();
}
for(ll i = N - 1;i >= 0;i--){
if(i + D < N) root1 -> update(i + D,mini[i + D]); //update (i+D)th position with value mini[i+D]
rightmost[i] = root1 -> bs(arr[i].first);
}
sort(arr,arr + N); //increasing A[i], if tie (same A[i]) from right to left
ll dp[N];
ll ans = 0;
for(auto u : arr){ //loop in order of increasing A[i], if tie (same A[i]) from right to left
ll ind = -1 * u.second;
dp[ind] = root2 -> query(ind) + 1;
root2 -> update(ind,rightmost[ind],dp[ind]); //future states can only come from ind if their original position <= rightmost[ind]
ans = max(ans,dp[ind]);
}
cout<<ans<<'\n';
}
