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.*;

public class Main {
    public static void main(String args[]) throws Exception {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        int len = s.length();
        int[] res = new int[len];

        int i = 0;
        while(i < len) {
        	int leftR = i;
        	while(i < len && s.charAt(i) == 'R') {
        		i++;
        	}
        	int rightR = i-1;

        	int leftL = i;
        	while(i < len && s.charAt(i) == 'L') {
        		i++;
        	}
        	int rightL = i - 1;
        	int cntR = rightR - leftR + 1;
        	int cntL = rightL - leftL + 1;
        	int cnt = (cntR + cntL) / 2;
        	res[rightR] = cnt;
        	res[leftL] = cnt;
        	if((cntR + cntL) % 2 != 0) {
        		if(cntR > cntL) {
        			if((cntR-1) % 2 == 0)
        				res[rightR]++;
        			else
        				res[leftL]++;
        		} else {
        			if((cntL-1) % 2 == 0)
        				res[leftL]++;
        			else
        				res[rightR]++;
        		}
        	}
        }
        for(int j = 0; j < len; j++)
        	if(j == len - 1)
        		System.out.println(res[j]);
        	else
        		System.out.print(res[j] + " ");
    }
}
