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) throws Exception {
        Scanner sc = new Scanner(System.in);
        String tmp_s = sc.next() + "L";
        String[] s = tmp_s.split("");
        int[] c = new int[s.length];
        int mass = 0;
        int left = 0;
        for(int i=0;i<s.length-1;i++){
            mass++;
            if(s[i].equals("R") && s[i+1].equals("L")){
                int tmp = (mass%2==0) ? mass/2 : (mass+1)/2;
                c[i] += tmp;
                c[i+1] += (mass-tmp);
                left = i+1;
                mass = 0;
                if(i+1 == s.length-1)c[i+1]=1;
            }
            if(s[i].equals("L") && s[i+1].equals("R")){
                int tmp = (mass%2==0) ? mass/2 : (mass+1)/2;
                c[left] += tmp;
                c[left-1] += (mass-tmp);
                mass = 0;
            }
            if(i+1 == s.length-1){
                int tmp = (mass%2==0) ? mass/2 : (mass+1)/2;
                c[left] += tmp;
                c[left-1] += (mass-tmp);
            }
        }
        for(int i=0;i<s.length-1;i++){
            System.out.print(c[i] + " ");
        }
    }
}