Given is a string S consisting of `L` and `R`.

Let N be the length of S. There are N squares arranged from left to right, and the i-th character of S from the left is written on the i-th square from the left.

The character written on the leftmost square is always `R`, and the character written on the rightmost square is always `L`.

Initially, one child is standing on each square.

Each child will perform the move below 10^{100} times:

* Move one square in the direction specified by the character written in the square on which the child is standing. `L` denotes left, and `R` denotes right.



Find the number of children standing on each square after the children performed the moves.

Constraints

* S is a string of length between 2 and 10^5 (inclusive).
* Each character of S is `L` or `R`.
* The first and last characters of S are `R` and `L`, respectively.

Input

Input is given from Standard Input in the following format:


S


Output

Print the number of children standing on each square after the children performed the moves, in order from left to right.

Examples

Input

RRLRL


Output

0 1 2 1 1


Input

RRLLLLRLRRLL


Output

0 3 3 0 0 0 1 1 0 2 2 0


Input

RRRLLRLLRRRLLLLL


Output

0 0 3 2 0 2 1 0 0 0 4 4 0 0 0 0
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    char[] s = scanner.next().toCharArray();
    int N = s.length;

    int[] res = new int[N];
    for (int i = 0; i < N - 1; i++) {
      if (s[i] == 'R' && s[i + 1] == 'L') {
        for (int j = 0; i - j >= 0 && s[i - j] == 'R'; j++)
          if (j % 2 == 0) res[i]++; else res[i + 1]++;
        for (int j = 0; i + 1 + j < N && s[i + 1 + j] == 'L'; j++)
          if (j % 2 == 0) res[i + 1]++; else res[i]++;
      }
    }
    StringBuilder sb = new StringBuilder();
    for (int i : res) sb.append(i).append(' ');
    sb.setLength(sb.length() - 1);
    System.out.println(sb.toString());
  }
}
