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.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Main {

	public static void main(String[] args) throws Exception {
		// TODO 自動生成されたメソッド・スタブ
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String S = in.readLine();

		long[] A = new long[S.length()];

		Arrays.fill(A, 0);

		long r=0;
		int rIndx = -1;
		long l=0;
		boolean flg = true;
		for(int i=0;i<S.length();i++){
			if(flg){
				if(S.charAt(i)=='R'){
					r++;
				}else{
					l=1;
					rIndx=i-1;
					flg=false;
				}
			}
			else{
				if(S.charAt(i)=='L'){
					l++;
				}else{
					A[rIndx]=(l+r)/2;
					A[rIndx+1]=(l+r)/2;

					if(Math.abs(l+r)%2==1){
						if(l%2==0){
							A[rIndx]++;
						}else{
							A[rIndx+1]++;
						}
					}
					flg=true;
					r=1;

				}

			}
		}
		A[rIndx]=(l+r)/2;
		A[rIndx+1]=(l+r)/2;

		if(Math.abs(l-r)%2==1){
			if(l%2==0){
				A[rIndx]++;
			}else{
				A[rIndx+1]++;
			}
		}
		for(int i=0;i<S.length()-1;i++){
			System.out.print(A[i]+" ");
		}
		System.out.println(A[S.length()-1]);
	}
}
