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.InputStreamReader;
import java.util.BitSet;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		String s = in.next();
		solve(s);
	}

	private static void solve(String s) {
		int lenR = 0;
		int lenL = 0;
		for (int i = 0; i < s.length(); i++) {
			if (s.charAt(i) == 'R')
				lenR++;
			else
				lenL++;
			if (i < s.length() - 1)
				if (s.charAt(i) == 'L' && s.charAt(i + 1) == 'R') {
					printSolve(lenR, lenL);
					lenR = 0;
					lenL = 0;
				}
		}
		printSolve(lenR, lenL);
		System.out.println();
	}

	private static void printSolve(int lenR, int lenL) {
		for (int i = 1; i < lenR; i++)
			System.out.print("0 ");
		System.out.print((lenR - (lenR / 2) + (lenL / 2)) + " " + ((lenR / 2) + lenL - (lenL / 2)) + " ");
		for (int i = 1; i < lenL; i++)
			System.out.print("0 ");
	}
}
