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[]){
    //入力
    Scanner scanner = new Scanner(System.in);
    String S = scanner.next();
    
    //初期値
    String[] arrayS = S.split("");
    int N = S.length();
	int[] nums = new int[N];
    for(int num: nums){
      num = 0;
    }
    
    int[] RLs = new int[N];
    RLs[0] = 0;
    int countRL = 1;
    
    int[] LRs = new int[N];
    LRs[0] = 0;
    int countLR = 1;
    
    //処理
    for(int i = 0; i < N - 1; i++){
      if("R".equals(arrayS[i]) &&  "L".equals(arrayS[i + 1])){
        //RLの場所
        RLs[countRL] = i;
        countRL++;
      }
      
      if("L".equals(arrayS[i]) && "R".equals(arrayS[i + 1])){
        //LRの場所
        LRs[countLR] = i;
        countLR++;
      }
    }
    
    LRs[countLR] = N - 1;
    
    for(int i = 1; i < N; i++){
      if(RLs[i] == 0 && i > 1){
        break;
      }
      //Lの個数
      int amountL = LRs[i] - RLs[i];
      
      if(amountL % 2 == 0){
        //Lの個数が偶数の場合の集まり方
        nums[RLs[i] + 1] += amountL / 2;
        nums[RLs[i]] += amountL / 2;
      } else {
        //Lの個数が奇数の場合の集まり方
        nums[RLs[i] + 1] += (amountL + 1) / 2;
        nums[RLs[i]] += (amountL - 1) / 2;
      }
      
      if(LRs[i] == 0){
        break;
      }
      
      //Rの個数
      int amountR = RLs[i] - LRs[i - 1];
      if(LRs[i - 1] == 0){
        amountR = RLs[i] - LRs[i - 1] + 1;
      }
      
      if(amountR % 2 == 0){
        //Rの個数が偶数の場合の集まり方
        nums[RLs[i]] += amountR / 2;
        nums[RLs[i] + 1] += amountR / 2;
      } else {
        //Rの個数が奇数の場合の集まり方
        nums[RLs[i]] += (amountR + 1) / 2;
        nums[RLs[i] + 1] += (amountR - 1) / 2;
      }
    }
    
    for(int num: nums){
      System.out.print(num);
      System.out.print(" ");
    }
  }
}
