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.*;
import java.util.Collections;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String S = sc.next();
        String[] s = S.split("");
        int N = s.length;
        long[] ans = new long[N];
        
        int rsum = 0;
        int lsum = 0;
        for(int i = 0; i < N; i++) {
            if(lsum == 0 && s[i].equals("R")) {
                rsum++;
                continue;
            }
            
            if(s[i].equals("L") && i < N-1) {
                lsum++;
                continue;
            }
            
            if(s[i].equals("R") || i == N-1) {
                if(i == N-1) {
                    lsum++;
                    i++;
                }
                if((rsum + lsum) % 2 == 0) {
                    ans[i - lsum - 1] = (rsum+lsum) / 2;
                    ans[i - lsum] = (rsum+lsum) / 2;
                }else {
                    if(rsum % 2 == 0) {
                        ans[i - lsum - 1] = (lsum+rsum+1) / 2 - 1;
                        ans[i - lsum] = (lsum+rsum+1) / 2;
                    }else {
                        ans[i - lsum - 1] = (lsum+rsum+1) / 2;
                        ans[i - lsum] = (lsum+rsum+1) / 2 - 1;
                    }
                }
                
                lsum = 0;
                rsum = 1;
            }
            
            
        }
        
        for(int i = 0; i < N; i++) {
            if(i < N - 1) {
                System.out.print(ans[i]+" ");
            }else {
                System.out.println(ans[i]);
            }
        }
    }
}