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);
        DGatheringChildren solver = new DGatheringChildren();
        solver.solve(1, in, out);
        out.close();
    }

    static class DGatheringChildren {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            // 入力
            char[] S = in.next().toCharArray();
            int n = S.length;
            int[] ans = new int[n];

            for (int i = 0; i < 2; i++) {
                int cnt = 0;
                for (int j = 0; j < n; j++) {
                    if (S[j] == 'R') {
                        cnt++;
                    } else {
                        ans[j] += cnt / 2;
                        ans[j - 1] += (cnt + 1) / 2;
                        cnt = 0;
                    }
                }
                reverse(S);
                reverse(ans);
                for (int j = 0; j < n; j++) {
                    if (S[j] == 'L') S[j] = 'R';
                    else S[j] = 'L';
                }
            }


            for (int i = 0; i < n; i++) {
                out.printf("%d ", ans[i]);
            }
            out.printf("\n");
        }

        void reverse(char[] array) {
            for (int i = 0; i < array.length / 2; i++) swap(array, i, array.length - i - 1);
        }

        void reverse(int[] array) {
            for (int i = 0; i < array.length / 2; i++) swap(array, i, array.length - i - 1);
        }

        void swap(int[] array, int i, int j) {
            int tmp = array[i];
            array[i] = array[j];
            array[j] = tmp;
        }

        void swap(char[] array, int i, int j) {
            char tmp = array[i];
            array[i] = array[j];
            array[j] = tmp;
        }

    }
}

