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 main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String a = sc.next();
		String s[] = new String[a.length()];
		int n[] = new int[a.length()];
		for(int i = 0; i < a.length(); i++){
			s[i] = a.substring(i, i+1);
		}
		int j;
		int cnt;
		boolean flg;
		for(int k = 0; k < a.length(); k++){
			flg = true;
			j = k;
			cnt = 0;
			if(s[k].equals("R")){
				while(true){
					if(s[j+1].equals("R")){
						j++;
						k++;
						cnt++;
						flg = !flg;
					}
					else{

							n[j] += cnt/2 + 1;
							n[j+1] += (cnt+1)/2;
						break;
					}
				}
			}
			else{
				while(j < s.length -1){
					if(s[j+1].equals("L")){
						j++;
						k++;
						cnt++;
				}
					else{
						break;
					}
				}

				while(true){
					if(s[j-1].equals("L")){
						j--;
						flg = !flg;

					}
					else{
							n[j] += cnt/2 + 1;
							n[j-1]+= (cnt+1)/2;
						break;
					}
				}

			}
		}
		for(int i = 0; i < n.length; i++){
			System.out.print(n[i] + " ");
		}

	}

}