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 in = new Scanner(System.in);
		String s = in.next();
		int[] ans = new int[s.length()];
		int a = 0;
		int b = 0;
		char now_c = 'R';
		int idx_r_to_l = 0;
		int i = 0;
		while(i<s.length()) {
			if(now_c=='R') {
				if(s.charAt(i)=='L') {
					idx_r_to_l = i-1;
					now_c = 'L';
				}
				if(i%2==0) a++;
				else b++;
			}else {
				if(s.charAt(i)=='R') {
					ans[idx_r_to_l] = idx_r_to_l%2==0?a:b;
					ans[idx_r_to_l+1] = (idx_r_to_l+1)%2==0?a:b;
					now_c = 'R';
					a = 0;
					b = 0;
				}
				if(i%2==0) a++;
				else b++;
			}
			i++;
		}
		ans[idx_r_to_l] = idx_r_to_l%2==0?a:b;
		ans[idx_r_to_l+1] = (idx_r_to_l+1)%2==0?a:b;
		
		for(int j=0;j<s.length();j++) {
			System.out.print(ans[j] + " ");
		}
		in.close();

	}

}
