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[] arg) {
    Scanner scan = new Scanner(System.in);
    String S = scan.next();
    int N = S.length();
    int A[] = new int[N];
    int Len = 0;
    int RL = 0;
    
    for (int i = 0; i < N; i++){
      if(S.substring(i,i+1).equals("R")){
        Len++;
        RL++;
      } else {
        if(i == N-1 || S.substring(i+1,i+2).equals("R")){
          Len++;
          if(Len%2 == 0){
            A[i-Len+RL]   = Len/2;
            A[i-Len+RL+1] = Len/2;
          } else {
            if(RL%2 == 0){
              A[i-Len+RL]   = Len/2;
              A[i-Len+RL+1]   = Len/2+1;
            } else {
              A[i-Len+RL]   = Len/2+1;
              A[i-Len+RL+1]   = Len/2;
            }
          }
          Len = 0;
          RL = 0;
        } else {
          Len++;
        }
      }
    }
    for (int i = 0; i < N; i++){
      System.out.print(A[i] + " ");
    }
  }
}