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.*;

public class Main {
  public static void main(String[] args) {
    Scanner scn = new Scanner(System.in);
    char[] S = (scn.nextLine() + "E").toCharArray();

    int[] answer = new int[S.length - 1];
    Arrays.fill(answer, 0);

    char prevChar = S[0];
    int prevBorderInd = 0;
    for (int i = 0; i < S.length; i++) {
      if (S[i] != prevChar) {
        if (S[i] == 'L') {
          if ((i - prevBorderInd) % 2 == 0) {
            answer[i] += (i - prevBorderInd) / 2;
            answer[i - 1] += (i - prevBorderInd) / 2;
          } else {
            answer[i] += (i - prevBorderInd) / 2;
            answer[i - 1] += (i - prevBorderInd) / 2 + 1;
          }
          prevChar = 'L';
        }

        if (S[i] == 'R' || S[i] == 'E') {
          if ((i - prevBorderInd) % 2 == 0) {
            answer[prevBorderInd] += (i - prevBorderInd) / 2;
            answer[prevBorderInd - 1] += (i - prevBorderInd) / 2;
          } else {
            answer[prevBorderInd] += (i - prevBorderInd) / 2 + 1;
            answer[prevBorderInd - 1] += (i - prevBorderInd) / 2;
          }
          prevChar = 'R';
        }
        prevBorderInd = i;

      }

    }
    for (int ans : answer) {
      System.out.printf("%d ", ans);
    }

  }
}