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) {
		Scanner sc = new Scanner(System.in);
		char[] S = sc.next().toCharArray();
		int N = S.length;
		int[] ans = new int[N];
		Arrays.fill(ans, 1);
		int[] cons = new int[N];
		Arrays.fill(cons, 1);
		for(int i = 1; i < N; i++) {
			if(S[i] == 'R' && S[i] == S[i-1]) {
				ans[i] += ans[i-1];
				ans[i-1] = 0;
				cons[i] = cons[i-1];
				cons[i]++;
			}
		}
		for(int i = N-2; i >= 0; i--) {
			if(S[i] == 'L' && S[i] == S[i+1]) {
				ans[i] += ans[i+1];
				ans[i+1] = 0;
				cons[i] = cons[i+1];
				cons[i]++;
			}
		}
		for(int i = 1; i < N; i++) {
			if(S[i] == 'L' && S[i] != S[i-1]) {
				int T = ans[i] + ans[i-1];
				if(cons[i-1] % 2 == 0 && cons[i] % 2 == 1) {
					ans[i-1] = T/2;
					ans[i] = (T+1)/2;
				}
				else {
					ans[i-1] = (T+1)/2;
					ans[i] = T/2;
				}
			}
		}
		for(int i = 0; i < N; i++)
			System.out.print(ans[i] + " ");
		System.out.println();

	}

}