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.InputStream;
import java.io.PrintStream;
import java.util.*;

public class Main {

  public static void main(String[] args) {
    solve(System.in, System.out);
  }

  static void solve(InputStream is, PrintStream os) {
    Scanner sc = new Scanner(is);

    /* read */
    String s = sc.next();
    int n = s.length();
    boolean[] dir = new boolean[n]; // left false, right true

    for (int i = 0; i < n; i++) {
      if (s.charAt(i) == 'R')
        dir[i] = true;
    }

    int[] children = new int[n];

    int l = 0;
    int r = 1;

    do {
      // R -> L
      while(r < n && dir[l] == dir[r]) r++;
      for (int i = l; i < r; i++) {
        if ((r - i) % 2 == 1) children[r - 1]++;
        else children[r]++;
      }
      if (r == n) break;
      // reset
      l = r;
      r = r + 1;
      // L -> R
      while(r < n && dir[l] == dir[r]) r++;
      for (int i = r - 1; l <= i ; i--) {
        if ((i - l) % 2 == 1) children[l - 1]++;
        else children[l]++;
      }
      l = r;
      r = r + 1;
    } while (r < n);
    
    for (int i = 0; i < n - 1; i++) {
      os.print(children[i] + " ");
    }
    os.println(children[n - 1]);
  }
}