//package ojuz;
import java.io.*;
import java.util.*;
import java.util.function.IntPredicate;
public class glo {
public static void main(String[] args) throws IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
String[] first = reader.readLine().split(" ");
int n = Integer.parseInt(first[0]), x = Integer.parseInt(first[1]);
int[] arr = Arrays.stream(reader.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
// longest lis ending at i
int[] endsAt = new int[n];
List<Integer> lis = new ArrayList<>();
for (int i = 0; i < n; i++) {
int finalI = i;
int pos = firstTrue(0, lis.size() - 1, j -> arr[finalI] <= lis.get(j));
endsAt[i] = pos + 1;
if (pos == lis.size()) lis.add(arr[i]);
else lis.set(pos, arr[i]);
}
lis.clear();
int best = 0;
for (int i = n - 1; i >= 0; i--) {
// since we need to find lis on a normal list, which is equal to finding lds on a reverse list
// which is equal to finding lis on the reversed list with the elements negated
int finalI = i;
// lis starting from the current (un-incremented) element
int pos = firstTrue(0, lis.size() - 1, j -> -arr[finalI] <= lis.get(j));
best = Math.max(best, pos + endsAt[i]);
// now build lis using incremented element
pos = firstTrue(0, lis.size() - 1, j -> (-arr[finalI] - x) <= lis.get(j));
if (pos == lis.size()) lis.add(-arr[i] - x);
else lis.set(pos, -arr[i] - x);
}
System.out.println(best);
}
public static int firstTrue(int lo, int hi, IntPredicate pred) {
hi++;
while (lo < hi) {
int mid = lo + (hi - lo) / 2;
if (pred.test(mid)) hi = mid;
else lo = mid + 1;
}
return lo;
}
}
