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.*;

public class Main{
	public static void main(String args[]) {
		
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		int N = s.length();
		int[] child = new int[N];
		Arrays.fill(child, 0);
		
		int[] R = new int[N];
		Arrays.fill(R, 1);
		int[] L = new int[N];
		Arrays.fill(L, 1);
		
		for(int i = 0; i < N - 2; i++) {
			if(s.charAt(i)=='R') {
				if(s.charAt(i + 1) == 'R') {
					if(s.charAt(i + 2) == 'R') {
						R[i + 2] += R[i];
						R[i] = 0;
					}
					else {
						child[i + 2] += R[i];
					}
				}
				else {
					child[i] += R[i];
				}
			}
			else {
				R[i] = 0;
			}
		}
		if(s.charAt(N - 2) == 'L') {
			R[N - 2] = 0;
		}
		else {
			child[N - 2] += R[N - 2];
		}
		
		for(int i = N - 1; i > 1; i--) {
			if(s.charAt(i)=='L') {
				if(s.charAt(i - 1) == 'L') {
					if(s.charAt(i - 2) == 'L') {
						L[i - 2] += L[i];
						L[i] = 0;
					}
					else {
						child[i - 2] += L[i];
					}
				}
				else {
					child[i] += L[i];
				}
			}
			else {
				L[i] = 0;
			}
		}
		if(s.charAt(1) == 'R')L[1] = 0;
		else {
			child[1] += L[1];
		}
		
		for(int i = 0; i < N; i++) {
			System.out.print(child[i]+((i != N-1)?" ":""));
		}
		System.out.println();
	}
}