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.io.PrintWriter;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        char[] s = sc.next().toCharArray();
        int n = s.length;
        sc.close();
        boolean flag = true;
        int[] ans = new int[n];

        int Lorigin = 0;
        int Rorigin = 0;
        for(int i=0;i<n;i++){
            if(s[i]=='R' && !flag){
               Rorigin = i;
               flag = true;
               ans[Lorigin-1] += (i-Lorigin)/2;
               ans[Lorigin] += (i-Lorigin+1)/2;
            }else if(s[i]=='L' && flag){
                Lorigin = i;
                flag = false;
                ans[i-1]+=(i-Rorigin+1)/2;
                ans[i]+=(i-Rorigin)/2;
            }

        }
        ans[Lorigin-1] += (n-Lorigin)/2;
        ans[Lorigin] += (n-Lorigin+1)/2;
        PrintWriter out = new PrintWriter(System.out);
        for(int i=0;i<n;i++){
            out.printf("%d ", ans[i]);
        }
        out.flush();

    }

}