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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        GatheringChildren solver = new GatheringChildren();
        solver.solve(1, in, out);
        out.close();
    }

    static class GatheringChildren {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            char[] s = in.next().toCharArray();
            int[] c = new int[s.length];

            int even = 0; // 偶数
            int odd = 0;

            for (int i = 0; i < s.length - 1; i++) {

                if ((i + 1) % 2 == 0) {
                    even++;
                } else {
                    odd++;
                }

                if (s[i] == 'R' && s[i + 1] == 'L') {
                    int r_pos = i + 1;
                    while (true) {
                        r_pos++;
                        if (s.length <= r_pos) {
                            break;
                        }
                        if (s[r_pos - 1] == 'L' && s[r_pos] == 'R') {
                            break;
                        }
                    }
                    for (int j = i + 1; j < r_pos; j++) {
                        if ((j + 1) % 2 == 0) {
                            even++;
                        } else {
                            odd++;
                        }
                    }

                    if (i % 2 == 0) {
                        c[i] = odd;
                        c[i + 1] = even;
                    } else {
                        c[i] = even;
                        c[i + 1] = odd;
                    }
                    i = r_pos - 1;
                    even = 0;
                    odd = 0;
                }
            }

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

    }
}

