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 sc = new Scanner(System.in);
    String S = sc.next();
    int[] A = new int[S.length()];
    int i = 0;
    while (i < S.length()) {
      int iRL = S.indexOf("RL", i);
      int numR = iRL - i;
      int iNextR = S.indexOf("R", iRL + 2);
      if (iNextR == -1) {
        iNextR = S.length();
      }
      int numL = iNextR - (iRL + 2);

      // RL(ab)
      int a = 1 + numR / 2 + (numL+1) / 2;
      int b = 1 + (numR + 1) / 2 + numL / 2;
      for (int j = i; j < iRL; j++) {
        A[j] = 0;
      }
      A[iRL] = a;
      A[iRL+1] = b;
        
      for (int j = iRL+2; j < iNextR; j++) {
        A[j] = 0;
      }
      
      // next
      i = iNextR;
    }
    for (int j = 0; j < A.length; j++) {
      System.out.print(A[j] + " ");
    }
  }  
}