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 debug(String str) {
        boolean debug = false;
        if (debug != false) {
            return;
        }
        System.out.println(str);
    }

    public static void main(String[] args) {

        String S;

        try (Scanner sc = new Scanner(System.in)) {
            S = sc.next();
        }
        long[] Math = new long[S.length()];
        for (int i = 0; i < Math.length; i++) {
            Math[i] = 0;
        }
        boolean LR[] = new boolean[S.length()];
        for (int i = 0; i < LR.length; i++) {
            if (S.charAt(i) == 'L') {
                LR[i] = true;
            } else {
                LR[i] = false;
            }
        }
        //Rの個数
        long countR = 0;
        for(int i = 0; i < LR.length; i++){
            if(LR[i] == false){
                countR++;
            }else{
                Math[i] += countR /2;
                Math[i-1] += countR /2;
                if(countR % 2 ==1){
                    Math[i-1] += 1;
                }
                countR =0;
            }
        }
        //Lの個数
        long countL = 0;
        for(int i = 0; i < LR.length; i++){
            if(LR[(LR.length - i) -1] == true){
                countL++;
            }else{
                Math[(LR.length - i) -1] += countL /2;
                Math[(LR.length - i) -1+1] += countL /2;
                if(countL % 2 ==1){
                    Math[(LR.length - i) -1+1] += 1;
                }
                countL =0;
            }
        }

        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < Math.length; i++) {
            long l = Math[i];
            sb.append(l).append(" ");
        }
        System.out.println(sb.toString());
        return;

    }
}
