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);
		char[] c = sc.next().toCharArray();
		int n = c.length;
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = 0;
		}
		int evn = 0;
		int odd = 0;
		int rtnPos = 0;
		for (int i = 0; i < n - 1; i++) {
			if (c[i] == 'R' && c[i + 1] == 'L') {
				a[i]++;
				if (i % 2 == 0) {
					a[i] += evn;
					a[i + 1] += odd;
				} else {
					a[i] += odd;
					a[i + 1] += evn;
				}
				evn = 0;
				odd = 0;
				rtnPos = i;
			} else if (c[i] == 'L' && c[i + 1] == 'R') {
				if (i % 2 == 0) {
					evn++;
				} else {
					odd++;
				}
				if (rtnPos % 2 == 0) {
					a[rtnPos] += evn;
					a[rtnPos + 1] += odd;
				} else {
					a[rtnPos] += odd;
					a[rtnPos + 1] += evn;
				}
				evn = 0;
				odd = 0;
			} else if (i % 2 == 0) {
				evn++;
			} else {
				odd++;
			}
		}

		if ((n - 1) % 2 == 0) {
			evn++;
		} else {
			odd++;
		}

		if (rtnPos % 2 == 0) {
			a[rtnPos] += evn;
			a[rtnPos + 1] += odd;
		} else {
			a[rtnPos] += odd;
			a[rtnPos + 1] += evn;
		}

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