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.PrintStream;
import java.util.Scanner;

public class Main {
    static final Scanner     sc = new Scanner(System.in);
    static final PrintStream so = System.out;

    public static void main(String[] args) {
        char[] s = nstr().toCharArray();
        int[] ans = new int[s.length];
        
        int i = 0;
        while(i < s.length) {
            int numR = 0;
            int numL = 0;
            while (s[i] == 'R') {
                i++;
                numR++;
            }
            while (i < s.length && s[i] == 'L') {
                i++;
                numL++;
            }
            
            int br = i-numL-1;
            int bl = br + 1;
            ans[br] = here(numR) + there(numL);
            ans[bl] = here(numL) + there(numR);
        }
        for (int i_ans = 0; i_ans < ans.length; i_ans++) {
            so.println(ans[i_ans]);
        }
    }

    private static int here(int i) {
        return (i+1)/2;
    }
    
    private static int there(int i) {
        return i/2;
    }

    private static long nlong() {
        return sc.nextLong();
    }

    private static int nint() {
        return sc.nextInt();
    }

    private static String nstr() {
        return sc.next();
    }

    private static char[] nsToChars() {
        return sc.next().toCharArray();
    }

    private static long[] nlongs(int n) {
        return nlongs(n, 0, 0);
    }

    private static int[] nints(int n) {
        return nints(n, 0, 0);
    }

    private static int[] nints(int n, int padL, int padR) {
        int[] a = new int[padL + n + padR];
        for (int i = 0; i < n; i++)
            a[padL + i] = nint();
        return a;
    }

    private static long[] nlongs(int n, int padL, int padR) {
        long[] a = new long[padL + n + padR];
        for (int i = 0; i < n; i++)
            a[padL + i] = nlong();
        return a;
    }

    private static String[] nstrs(int n) {
        String[] a = new String[n];
        for (int i = 0; i < n; i++)
            a[i] = nstr();
        return a;
    }

    private static char[][] nsToChars2D(int h, int w) {
        return nsToChars2D(h, w, 0);
    }

    private static char[][] nsToChars2D(int h, int w, int pad) {
        char[][] a2 = new char[h + pad * 2][w + pad * 2];
        for (int i = 0; i < h; i++)
            System.arraycopy(nsToChars(), 0, a2[pad + i], pad, w);
        return a2;
    }
}
