#include <bits/stdc++.h>
using namespace std;
const int maxN = 2e5 + 5;
int N, X, T[maxN];
namespace subtask1{
bool check(){
return X <= 50 && N <= 50;
}
int b[55], f[55];
int res = 0;
void sol(int a[]){
for(int i = 1; i <= N; ++i)f[i] = 0;
int ans = 0;
for(int i = 1; i <= N; ++i){
f[i] = 1;
for(int j = 1; j < i; ++j){
if(b[j] < b[i]) f[i] = max(f[i], f[j] + 1);
}
ans = max(ans, f[i]);
}
res = max(res, ans);
}
void solve(){
for(int d = -X; d <= X; ++d){
for(int i = 1; i <= N; ++i){
for(int j = i; j <= N; ++j){
for(int k = 1; k <= N; ++k)b[k] = T[k];
for(int x = i; x <= j; ++x)b[x] += d;
sol(b);
}
}
}
cout << res;
}
}
namespace subtask2{
bool check(){
return N <= 1000;
}
void solve(){
}
}
namespace subtask3{
int it[4 * maxN];
int comp[maxN];
bool check(){
return X == 0;
}
void update(int id, int l, int r, int p, int v){
if(l > p || r < p) return ;
if(l == r){
it[id] = v;
return ;
}
int mid = l + r >> 1;
if(p <= mid) update(id << 1, l, mid, p, v);
else update(id << 1 | 1, mid + 1, r, p, v);
it[id] = max(it[id << 1], it[id << 1 | 1]);
}
int get(int id, int l, int r, int u, int v){
if(l > v || r < u) return 0;
if(l >= u && r <= v) return it[id];
int mid = l + r >> 1;
return max(get(id << 1, l, mid, u, v), get(id << 1 | 1, mid + 1, r, u, v));
}
void solve(){
for(int i = 1; i <= N; ++i)comp[i] = T[i];
sort(comp + 1, comp + 1 + N);
for(int i = 1; i <= N; ++i){
T[i] = lower_bound(comp + 1, comp + 1 + N, T[i]) - comp;
update(1, 1, N, T[i], get(1, 1, N, 1, T[i] - 1) + 1);
}
cout << it[1];
}
}
namespace subtask4{
bool check(){
return X <= 5 && N <= 50000;
}
void solve(){
}
}
namespace subtask5{
bool check(){
return X == 1000000000;
}
void solve(){
}
}
namespace subtask6{
void solve(){
}
}
int main(){
ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
cin >> N >> X;
for(int i = 1; i <= N; ++i)cin >> T[i];
if(subtask1 :: check()) return subtask1 :: solve(), 0;
if(subtask2 :: check()) return subtask2 :: solve(), 0;
if(subtask3 :: check()) return subtask3 :: solve(), 0;
if(subtask4 :: check()) return subtask4 :: solve(), 0;
if(subtask5 :: check()) return subtask5 :: solve(), 0;
return subtask1 :: solve(), 0;
return 0;
}
/*
8 0
7 3 5 12 2 7 3 4
*/
