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 s=sc.next();
		int[] ans=new int[s.length()];
		int count=1;
		int in=0;
		for(int i=0; i<s.length()-1; i++){
			if(s.charAt(i)=='L' && s.charAt(i+1)=='R'){
				ans[in]=count;
				count=1;
				continue;
			}
			if(s.charAt(i)=='R' && s.charAt(i+1)=='L'){
				ans[i]=count;
				in=i+1;
				count=1;
			}else{
				count++;
			}
		}
		ans[in]=count;
		for(int i=0; i<s.length()-1; i++){
			if(s.charAt(i)=='R' && s.charAt(i+1)=='L'){
				if(ans[i]!=ans[i+1]){
					int t=(ans[i]+ans[i+1])/2;
					int y=ans[i]+ans[i+1]-t;
					if(ans[i]>ans[i+1]){
						if(ans[i]%2==0){
							ans[i]=t;
							ans[i+1]=y;
						}else{
							ans[i]=y;
							ans[i+1]=t;
						}
					}else{
						if(ans[i+1]%2==0){
							ans[i+1]=t;
							ans[i]=y;
						}else{
							ans[i+1]=y;
							ans[i]=t;
						}
					}
				}
			}
			System.out.print(ans[i]+" ");
		}
		System.out.print(ans[s.length()-1]);
	}
}
