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 scanner = new Scanner(System.in);
        char[] s = scanner.next().toCharArray();
        int[] children = new int[s.length];

        for (int i = 0; i < s.length; i++) {
            children[i] = 0;
        }

        int j = 0;
        while (s[++j] != 'L');

        for (int i = 0; i < s.length; i++) {
            if (s[i] == 'R') {
                if (0 < i && s[i - 1] != 'R') {
                    for (j = i + 1; j < s.length; j++) {
                        if (s[j] == 'L')
                            break;
                    }
                }
                if ((j - i) % 2 == 1) {
                    children[j - 1]++;
                } else {
                    children[j]++;
                }
            } else {
                if (s[i - 1] != 'L') {
                    for (j = i - 1; j >= 0; j--) {
                        if (s[j] == 'R')
                            break;
                    }
                }
                if ((i - j) % 2 == 1) {
                    children[j + 1]++;
                } else {
                    children[j]++;
                }
            }
        }

        for (int i = 0; i < s.length; i++) {
            System.out.print(children[i] + " ");
        }
    }
}

