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 stdIn=new Scanner(System.in);
		String S=stdIn.next();
		int N=S.length();
		int roop[]=new int[S.length()];
		char A[]=new char[N];
		int ans[]=new int[N];
		int z=0,y=0,x=0;
		while(z<N){
			A[z]=S.charAt(z);
			if(z!=0)
				if(A[z]=='L'&&A[z-1]=='R'){
					roop[y]=z-1;
					roop[y+1]=z;
					y+=2;
				}
			z++;
		}z=0;
		while(z<N){
			if(z==roop[x+1]){
				ans[roop[x+1]]++;
				x+=2;
			}
			else{
				if(z<roop[x]&&A[z]=='R')
					if((roop[x]-z)%2==1)
						ans[roop[x+1]]++;
					else
						ans[roop[x]]++;
				if(x!=0&&z>roop[x-1]&&A[z]=='L')
					if((z-roop[x-1])%2==1)
						ans[roop[x-2]]++;
					else
						ans[roop[x-1]]++;
				if(z==roop[x])
					ans[roop[x]]++;
			}
			z++;
		}z=0;
		while(z<N){
			System.out.println(ans[z]);
			z++;
		}
	}
}