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();
        char str[] = s.toCharArray();
//        char str[] = new char[s.length()];
//        for (int i=0; i<s.length(); i++){
//            str[i] = s.indexOf
//        }

        int i = 0;
        while(i<s.length()){
            int r = 0;
            int l = 0;

            while(i<s.length() && str[i]=='R'){
                r++;
                i++;
            }
            while(i<s.length() && str[i]=='L'){
                l++;
                i++;
            }

            int R;
            int L;

            R = (int)Math.ceil(r/2.0);
            R += (int)l/2.0;

            L = (int)Math.ceil(l/2.0);
            L += (int)r/2.0;

            for (int j=0; j<r-1; j++) System.out.print(0+" ");
            System.out.print(R+" ");
            System.out.print(L+" ");
            for (int j=0; j<l-1; j++) System.out.print(0+" ");

        }

        }

    }
