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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.stream.Stream;
import java.util.TreeSet;

public class Main {
    public static void main(String[] args) throws Exception {
        
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String S = br.readLine();
        int N = S.length();
        
        int[] childrenNum = new int[N];
        TreeSet<Integer> RLnoRnoIndex = new TreeSet<>();
        
        for(int i = 0; i < N - 1; i++){
            if ( S.charAt(i) == 'R' && S.charAt(i+1) == 'L' ) {
                RLnoRnoIndex.add(i);
            }
        }
        
        //Rについて最初からループ
        for(int i = 0; i < N - 1; i++){
            if ( S.charAt(i) == 'R' && S.charAt(i+1) == 'R' ) {
                int RIndex = RLnoRnoIndex.higher(i);
                if ( (RIndex - i) % 2 == 0){
                    childrenNum[RIndex]++;
                } else {
                    childrenNum[RIndex+1]++;
                }
            }
            
            if ( S.charAt(i) == 'R' && S.charAt(i+1) == 'L' ) {
                childrenNum[i]++;
            }
        }
        
        //Lについて最後からループ
        for(int i = N - 1; i >= 1; i--){
            if ( S.charAt(i) == 'L' && S.charAt(i-1) == 'L' ) {
                int RIndex = RLnoRnoIndex.lower(i);
                if ( (i - RIndex) % 2 == 0 ){
                    childrenNum[RIndex]++;
                } else {
                    childrenNum[RIndex+1]++;
                }
            }
            
            if ( S.charAt(i) == 'L' && S.charAt(i-1) == 'R' ) {
                childrenNum[i]++;
            }
        }
        
        StringBuilder sb = new StringBuilder();
        
        for(int i = 0; i < N; i++){
            sb.append(String.valueOf(childrenNum[i]));
            if ( i != N - 1 ) sb.append(" ");
        }
        
        System.out.println(sb);
    }
}
