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.*;
import static java.lang.Math.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		// 入力
		String s = sc.next();
		
		// 計算
		int[] result = new int[s.length()];
		char c = 'R';
		int cnt = 1;
		for(int i = 0; i < s.length(); i++){
		    if(i == 0){
		        c = s.charAt(i);
		    }else if(c == s.charAt(i)){
		        cnt++;
		    }else{
		        result[i-1] += (cnt+1)/2;
		        result[i] += cnt/2;
		        cnt = 0;
		    }
		}
		cnt = 1;
		for(int i = s.length()-1; i >= 0; i--){
		    if(i == s.length()-1){
		        c = s.charAt(i);
		    }else if(c == s.charAt(i)){
		        cnt++;
		    }else{
		        result[i+1] += (cnt+1)/2;
		        result[i] += cnt/2;
		        cnt = 0;
		    }
		}
		
		// 出力
		for(int i = 0; i < s.length(); i++){
		    System.out.print(result[i] + " ");
		}
	}
}
