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.InputStream;
import java.io.PrintStream;
import java.util.Arrays;
import java.util.Scanner;

import static java.lang.Math.ceil;
import static java.lang.Math.floor;

public class Main {
    public static void main(String[] args) throws Exception {
        solve(System.in, System.out);
    }

    static void solve(InputStream is, PrintStream os) {
        // Your code here!
        Scanner scan = new Scanner(is);
        String S = scan.next();
        int N = S.length();
        char arrayS[] = S.toCharArray();
        int ans[] = new int[N];
        Arrays.fill(ans, 0);

        char prevS = 'R';
        int count = 0;

        for(int i = 0; i < N; i++) {
            if(prevS == 'R' && arrayS[i] == 'L') {
                ans[i-1] = (int) ceil(count / 2.0);
                ans[i] = (int) floor(count / 2.0);
                count = 0;
            }
            else if(arrayS[i] == 'R')
                count++;

            prevS = arrayS[i];
        }

        prevS = 'L';
        count = 0;
        for(int i = N - 1; i >= 0; i--) {
            if(prevS == 'L' && arrayS[i] == 'R') {
                ans[i+1] += ceil(count / 2.0);
                ans[i] += floor(count / 2.0);
                count = 0;
            }
            else if(arrayS[i] == 'L')
                count++;

            prevS = arrayS[i];
        }

        for(int i = 0; ; ) {
            os.print(ans[i]);

            if(++i < N)
                os.print(" ");
            else {
                os.println("");
                return;
            }
        }
    }
}