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 s = sc.nextLine();
		int[] a = new int[s.length()];

		int current = 0;
		for (int i = 0; i < s.length(); i++) {
			if (s.charAt(i) == 'R') {
				current++;
			} else if (current > 0) {
				a[i] += current / 2;
				a[i - 1] += (current - 1) / 2 + 1;
				current = 0;
			}
		}

		current = 0;
		for (int i = s.length() - 1; i >= 0; i--) {
			if (s.charAt(i) == 'L') {
				current++;
			} else if (current > 0) {
				a[i] += current / 2;
				a[i + 1] += (current - 1) / 2 + 1;
				current = 0;
			}
		}

		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < a.length; i++) {
			sb.append(a[i]);
			sb.append(" ");
		}
		System.out.println(sb.toString());
	}
}
