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;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        char[] c = sc.next().toCharArray();
        int[] cnt = new int[c.length];
        
        for (int i = 0; i < c.length;) {
            int index = i;
            int tmp_cnt = 0;
            char tmp_c = c[i];
            while (tmp_c == c[i]) {
                tmp_cnt++;
                i++;
                if (i >= c.length)
                    break;
            }

            if (tmp_c == 'R') {
                if (tmp_cnt % 2 == 1) {
                    cnt[i-1]++;
                }
                cnt[i] += tmp_cnt / 2;
                cnt[i-1] += tmp_cnt / 2;
            } else {
                if (tmp_cnt % 2 == 1) {
                    cnt[index]++;
                }
                cnt[index] += tmp_cnt / 2;
                cnt[index-1] += tmp_cnt / 2;
            }
        }

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