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.Arrays;
import java.util.Scanner;
import java.util.stream.Collectors;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        int[] a = new int[s.length()];
        Arrays.fill(a, -1);
        for (int i = 0; i < s.length() - 1; i++) {
            if (s.charAt(i) == 'R' && s.charAt(i + 1) == 'L') {
                a[i] = 1;
                a[i + 1] = 1;
            }
        }
        int cnt = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == 'R' && a[i] == -1) {
                a[i] = 0;
                cnt++;
            } else if (a[i] != -1 && cnt > 0) {
                if (cnt % 2 == 0) {
                    a[i] += cnt / 2;
                    a[i + 1] += cnt / 2;
                } else {
                    a[i] += cnt / 2;
                    a[i + 1] += cnt / 2 + 1;
                }
                cnt = 0;
            }
        }
        cnt = 0;
        for (int i = s.length() - 1; i >= 0; i--) {
            if (s.charAt(i) == 'L' && a[i] == -1) {
                a[i] = 0;
                cnt++;
            } else if (a[i] != -1 && cnt > 0) {
                if (cnt % 2 == 0) {
                    a[i] += cnt / 2;
                    a[i - 1] += cnt / 2;
                } else {
                    a[i] += cnt / 2;
                    a[i - 1] += cnt / 2 + 1;
                }
                cnt = 0;
            }
        }
        System.out.println(Arrays.stream(a).mapToObj(String::valueOf).collect(Collectors.joining(" ")));
    }
}