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();
    char[] s = S.toCharArray();
    int len = S.length();
    int[] ans = new int[len];
    for(int i = 0;i < len - 1;i++){
      if(s[i] == 'R' && s[i+1] =='L'){
        int rp = 0;
        int lp = 0;
        if(i >= 1){
          for(int x = i - 1;x >= 0;x--){
            if(s[x] == 'R'){
              rp++;
            }else{
              break;
            }
          }
        }
        ans[i] += rp / 2;
        ans[i+1] += rp - (rp / 2);
        if(i + 2 < len){
          for(int y = i + 2;y < len;y++){
            if(s[y] == 'L'){
              lp++;
            }else{
              break;
            }
          } 
        }
        ans[i] += lp - (lp / 2);
        ans[i+1] += lp / 2;
        ans[i]++;
        ans[i+1]++;
      }
    }
    for(int z = 0;z < len;z++){
      System.out.print(ans[z] + " " );
    }
  }
}