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.*;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);

		// 入力
		char[] s = (sc.next()).toCharArray();
		int[] result = new int[s.length];
		float r = 1;
		int l_i = 0;
		float l = 0;
		for(int i = 0; i < s.length; i++){
		    if(i > 0 && s[i - 1] == 'R' && s[i] == 'R'){
		        result[i - 1] = 0;
		        r = r + 1;
		    }else if(i > 0 && s[i - 1] == 'R' && s[i] == 'L'){
		        result[i - 1] = (int)ceil(r/2);
		        result[i] = 1 + (int)floor(r/2);
		        r = 0;
		        l_i = i;
		    }else if(i > 0 && s[i - 1] == 'L' && s[i] == 'R'){
		        result[l_i - 1] = result[l_i - 1] + (int)ceil(l/2);
		        result[l_i] = result[l_i] + (int)floor(l/2);
		        r = 1;
		        l = 0;
		    }else if(i > 0 && s[i - 1] == 'L' && s[i] == 'L'){
		        result[i] = 0;
		        l = l + 1;
		    }
		}
		result[l_i - 1] = result[l_i - 1] + (int)ceil(l/2);
		result[l_i] = result[l_i] + (int)floor(l/2);
		
		// 計算

		// 出力
		for(int i = 0; i < result.length; i++){
            System.out.print(result[i] + " ");
		}
		
	}
}
