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);
        String S = sc.next();

        int n=S.length();
        char[] s = S.toCharArray();
        int[] ans = new int[n];

        int i=0;
        while(i<n){
            int Rcnt=0;
            while(i<n && s[i]=='R'){
                i++;
                Rcnt++;
            }
            int ni=i;
            ans[ni-1]=ans[ni-1]+(Rcnt+1)/2;
            ans[ni]=ans[ni]+Rcnt/2;

            int Lcnt=0;
            while(i<n && s[i]=='L'){
                i++;
                Lcnt++;
            }
            ans[ni-1]=ans[ni-1]+Lcnt/2;
            ans[ni]=ans[ni]+(Lcnt+1)/2;
        }
        for (int j = 0; j < n; j++) {
            System.out.print(ans[j] + " ");
        }
    }
}
