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);
		StringBuilder sb = new StringBuilder();
		char[] s = sc.next().toCharArray();
		int r = 0, l;
		int[] a = new int[s.length];
		for (int i = 0; i < s.length - 1; i++) {
			if (s[i] == 'R' && s[i + 1] == 'L') {
				a[i]++;
				a[i + 1]++;

				for (int j = r; j < i; j++) {
					if ((i - j) % 2 == 0) {
						a[i]++;
					} else {
						a[i + 1]++;
					}
				}

				for (int j = i + 2; j < s.length; j++) {
					if (s[j] == 'L') {
						if ((j - i) % 2 == 0) {
							a[i]++;
						} else {
							a[i + 1]++;
						}
					} else {
						i = j - 1;
						r = i + 1;
						break;
					}
				}
			}
		}
		sb.append(a[0]);
		for (int i = 1; i < s.length; i++) {
			sb.append(" " + a[i]);
		}
		System.out.println(sb);
	}
}