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.*;
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];
    Arrays.fill(a,0);
    int f=0;
    int b;
    int t;
    boolean flag=true;
    while(flag){
      b=s.indexOf("RL",f);
      t=s.indexOf("LR",b);
      if(t==-1){
        flag=false;
        t=n-1;
      }
      a[b]=a[b]+(b-f+1)/2+(b-f+1)%2+(t-b)/2;
      a[b+1]=a[b+1]+(b-f+1)/2+(t-b)/2+(t-b)%2;
      f=t+1;
    }
    for(int i=0;i<n;i++){
      System.out.println(a[i]);
    }
  }
}