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){
		try(Scanner sc = new Scanner(System.in)) {

			String s = sc.next();
			
			//R...RL...Lのブロックに分離してそれぞれごとに集計すればよい
			//RRRLLの場合(start = 0, sep = 3, len=5)
			//00320
			//RRRRLの場合(start = 0, sep = 4, len=5)
			//00023
			//RRRLの場合 (start = 0, sep = 3, len=4)
			//0022
			
			char L = 'L';
			char R = 'R';
			
			char[] sArray = (s + "T").toCharArray();
			int[] result = new int[s.length()];
			
			int start = 0;
			int sep = 0;
			boolean isRight = true;
			for(int i = 0 ; i < sArray.length ; i++ ) {
				if ( isRight ) {
					if ( sArray[i] == R ) {
						//continue
					}
					else {
						//divider
						isRight = false;
						sep = i;
						continue;
					}
				}
				else {
					if ( sArray[i] == L ) {
						//continue
					}
					else {
						//termination, so count it
						int blockLen = i - start;
						
						if ( blockLen % 2 == 0 ) {
							result[sep-1] = blockLen / 2;
							result[sep] = blockLen / 2;
						}
						else {
							int gap = sep - start;
							result[sep-1] = (blockLen / 2) + gap % 2;
							result[sep] = (blockLen / 2) + (gap+1) % 2;
						}
						
						start = i;
						isRight = true;
					}
				}
			}
			
			for(int i = 0 ; i < result.length ; i++ ) {
				System.out.print(result[i] + " ");
			}
		}
	}
}