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.io.PrintWriter;
import java.util.Scanner;

public class Main{

	public static void main(String[] args){
		//入力
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		sc.close();

		//処理
		int n = s.length();
		int[] ans = new int[n];
		for(int i = 0; i < n; i++) ans[i] = 0;
		int last = 0;
		for(int i = 0; i < n; i++){
			if(s.charAt(i) == 'R'){
				last++;
			}else{
				ans[i] += last/2;
				ans[i-1] += (last+1)/2;
				last = 0;
			}
		}
		last = 0;
		for(int i = n-1; i >= 0; i--){
			if(s.charAt(i) == 'L'){
				last++;
			}else{
				ans[i] += last/2;
				ans[i+1] += (last+1)/2;
				last = 0;
			}
		}
		
		//出力
		PrintWriter out = new PrintWriter(System.out);
		for(int i = 0; i < n; i++){
			out.print(ans[i]);
			if(i != n-1) out.print(" ");
			else out.println();
		}
		out.flush();
	}
}