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);
		String str = sc.next();
		int len = str.length();
		char[] rl = str.toCharArray();
		int[] count_r = new int[len], count_l=new int[len], count = new int[len];
		int num=0;
		for(int i=len-1;i>=0;i--) {
			if(rl[i]=='L') num=0;
			else {
				num++;
				count_r[i]=num;
			}
		}
		for(int i=0;i<len;i++) {
			if(rl[i]=='R') num=0;
			else {
				num++;
				count_l[i]=num;
			}
		}
		for(int i=0;i<len;i++) {
			if(count_r[i]!=0) {
				if(count_r[i]%2==0) count[i+count_r[i]]++;
				else count[i+count_r[i]-1]++;
			}else {
				if(count_l[i]%2==0) count[i-count_l[i]]++;
				else count[i-count_l[i]+1]++;
			}
		}
		for(int i=0;i<len;i++) System.out.print(count[i]+" ");
		System.out.println();
	}
}