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.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        char[]s=sc.next().toCharArray();
        int n=s.length;
        int c[]=new int[n];
        char a='R';
        ArrayList<Integer>aaa=new ArrayList<Integer>();
        aaa.add(0);
        int ind=0;
        for(int i=0;i<n;i++){
            if(a==s[i]){
                ind++;
            }
            else{
                a=(a=='R')?'L':'R';
                aaa.add(ind);
                ind++;
            }
        }
        if(aaa.size()%2==0)aaa.add(ind);
        for(int i=0;i<aaa.size()/2;i++){
            int x2=aaa.get(2*i+1);
            int y=aaa.get(2*i+2)-x2;
            int x=x2-aaa.get(2*i);
            if(x%2==0){
                if(y%2==0){
                    c[x2-1]=(x+y)/2;
                    c[x2]=(x+y)/2;
                }else{
                    c[x2-1]=(x+y)/2;
                    c[x2]=(x+y)/2+1;
                }
            }
            else{
                if(y%2==0){
                    
                    c[x2-1]=(x+y)/2+1;
                    c[x2]=(x+y)/2;
                }else{
                    c[x2-1]=(x+y)/2;
                    c[x2]=(x+y)/2;
                }
            }
        }
        for(int i=0;i<n;i++)
        System.out.print(c[i]+" ");
    }
}