Sort million 32-bit integers

MillionDWords.java

import java.util.Random;
import java.util.Arrays;
import java.time.*;
import java.time.temporal.*;

public class MillionDWords {
  final static int million = 1_000_000;
  final static int runs = 100;
  public static void main(String []argc) {
    int[] numbers = new Random().ints(million, Integer.MIN_VALUE, Integer.MAX_VALUE).toArray();
    int[] numbers2 = Arrays.copyOf(numbers, numbers.length);

    LocalTime initTime = LocalTime.now();
    Arrays.sort(numbers);
    System.out.printf("Arrays.sort() sorted in %10d ns\n",
        ChronoUnit.NANOS.between(initTime, LocalTime.now()));
    // System.out.println(Arrays.toString(numbers));

    initTime = LocalTime.now();
    LSDSort(numbers2);
    System.out.printf("LSD() sorted in         %10d ns\n",
        ChronoUnit.NANOS.between(initTime, LocalTime.now()));
    // System.out.println(Arrays.toString(numbers2));

    System.out.printf("Arrays are equals: %b\n", 
        Arrays.equals(numbers, numbers2));
  }

  private static int byteAt(int number, int position) {
    return ((number ^ (1 << (7+8+8+8))) >>> (8 * (3-position))) & 0xFF;
  }

  private static void LSDSort(int[] a) {
    final int N = a.length;
    final int R = 1<<8;
    final int W = 4;
    int[] aux = new int[N];

    for (int d = W-1; d >= 0; d--) {
      int[] count = new int[R+1];
      for (int i = 0; i < N; i++)
        count[byteAt(a[i], d) + 1]++;

      for (int r = 0; r < R; r++)
        count[r+1] += count[r];

      for (int i = 0; i < N; i++)
        aux[count[byteAt(a[i], d)]++] = a[i];

      for (int i = 0; i < N; i++)
        a[i] = aux[i];
    }

  }
}