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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String str[] = br.readLine().split("");

		int n = str.length;

		int a[] = new int[n];

		for (int i = 0; i < n;) {
			int r = 0;
			int l = 0;
			int now = 0;

			while (i < n && str[i].equals("R")) {
				i++;
				r++;
			}

			now = i;
			a[now - 1] = a[now - 1] + (r + 1) / 2;
			a[now] = a[now] + r / 2;

			while (i < n && str[i].equals("L")) {
				i++;
				l++;
			}
			a[now - 1] = a[now - 1] + l / 2;
			a[now] = a[now] + (l + 1) / 2;
		}

		System.out.print(a[0]);
		for (int j = 1; j < n; j++) {
			System.out.print(" " + a[j]);
		}
		System.out.println();
	}
}
