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.*;

class Main{
    public static void main(String[] args){
        
        Scanner sc = new Scanner(System.in);
        
        String S = sc.next();
        int l = S.length();
        char[] a = S.toCharArray();
        int child = 0;
        int[] ans = new int[l];
        int x = 0;
        int y = 0;

        for(int i=0; i<l-1; i++){
            child += 1;
            if(a[i]!=a[i+1] && a[i]=='R'){
                x = i;
                y = i + 1;
                ans[i + 1] += child / 2;
                ans[i] += child - child / 2 ;
                child = 0;
            }
            else if(a[i]!=a[i+1] && a[i]=='L'){
                ans[x] += child / 2;
                ans[y] += child - child / 2;
                child = 0;
            }
        }
        child += 1;
        ans[x] += child / 2;
        ans[y] += child - child / 2;
        System.out.print(ans[0]);
        for(int i=1; i<l-1; i++){
            System.out.print(" " + ans[i]);
        }
        System.out.println(" " + ans[l - 1]);
    }
}