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.*;
class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.next();
        int x, y = 0;
        for(int i = 0; i<str.length(); i++) {
            x = -1;
            y = -1;
            boolean isR = true;
            while(i<str.length()) {
                char c = str.charAt(i);
                if(c=='R' && !isR) {
                    i--;
                    break;
                }
                if(c=='L') {
                    isR = false;
                    y++;
                }
                if(c=='R') {
                    x++;
                }
                i++;
            }
            for(int j = 0; j<x; j++) System.out.print("0 ");
            System.out.print((1+x/2+(int)Math.ceil((double)y/(double)2))+" "+(1+y/2+(int)Math.ceil((double)x/(double)2))+" ");
            for(int j = 0; j<y; j++) System.out.print("0 ");
        }
        System.out.println();

    }
}