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.Scanner;
import java.util.StringJoiner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        int slen = s.length();
        int[][] rl = new int[slen][2];

        int i = 0;
        int j = 0;
        while (i < slen) {
            while (i < slen && s.charAt(i) == 'R') {
                rl[j][0]++;
                i++;
            }
            while (i < slen && s.charAt(i) == 'L') {
                rl[j][1]++;
                i++;
            }
            j++;
        }

        int[] ans = new int[slen];
        int m = 0;
        for (int k = 0; k < slen - 1; k++) {
            if (s.charAt(k) != 'R' || s.charAt(k + 1) != 'L') {
                ans[k] = 0;
                continue;
            }
            ans[k] = (rl[m][0] + 1) / 2 + rl[m][1] / 2;
            ans[k + 1] = rl[m][0] / 2 + (rl[m][1] + 1) / 2;
            m++;
            k++;
        }

        StringJoiner sj = new StringJoiner(" ");
        for (int k = 0; k < ans.length; k++) {
            sj.add(String.valueOf(ans[k]));
        }
        System.out.println(sj.toString());
    }

}
