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[] n = new int[s.length()];
		for(int i=0;i<s.length();i++) {
			n[i] =0;
		}
		int j=Integer.MIN_VALUE,k=Integer.MAX_VALUE;
		for(int i=0;i<s.length();i++) {
			if(s.charAt(i)=='R') {
				if(j<i) {
					k=Integer.MAX_VALUE;
					for(j=i;j<s.length();j++) {
						if(s.charAt(j)=='L') {
							break;
						}
					}
				}
				if(((j-i)%2) == 1) {
					n[j-1]++;
				}else {
					n[j]++;
				}
			}else {
				if(k>i) {
					for(k=i;k>=0;k--) {
						if(s.charAt(k)=='R') {
							break;
						}
					}
				}
				
				if(((i-k)%2) == 1) {
					n[k+1]++;
				}else {
					n[k]++;
				}
			}
		}
		for(int i=0;i<n.length;i++) {
			System.out.print(n[i] + " ");
		}
		System.out.println();
		
	}
}