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) {
    Scanner sc = new Scanner(System.in);
    String s = sc.next();
    int n = s.length();
    int[] a = new int[n];
    ArrayList<Integer> list = new ArrayList<Integer>();
    int[] ans = new int[n];
    list.add(0);
    for(int i = 0; i < n; i++) {
      if(s.charAt(i) == 'R') {
        
      } else {
        a[i] = 1;
      }
      if((i < n - 1) && (s.charAt(i) != s.charAt(i + 1))) list.add(i + 1);
    }
    list.add(n);
    int[] b = new int[n];
    for(int i = 0; i < list.size() - 1; i++) {
      for(int j = list.get(i); j < list.get(i + 1); j++) {
        if((i % 2) == 0) {
          b[j] = list.get(i + 1);
        } else {
          b[j] = list.get(i) - 1;
        }
      }
    }
    for(int i = 0; i < n; i++) {
      if(a[i] == 0) {
        int t = b[i] - i;
        if((t % 2) == 0) {
          ans[b[i]]++;
        } else {
          ans[b[i] - 1]++;
        }
      } else {
        int t = i - b[i];
        if((t % 2) == 0) {
          ans[b[i]]++;
        } else {
          ans[b[i] + 1]++;
        }        
      }
    }
    for(int i = 0; i < n; i++) {
      System.out.println(ans[i]);
    }
  }
}