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.PrintWriter;
import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		char[] S = sc.next().toCharArray();
		int N = S.length;
		
		int index = 0;
		while(index < N) {
			int[] LRcount = getLR(S, index);
			int[] ans = getChild(LRcount);
			for(int a : ans) {
				pw.println(a);
			}
			index += LRcount[0]+LRcount[1];
		}
		
		pw.close();
		sc.close();
	}
	
	static int[] getLR(char[] S, int index) {
		int[] count = new int[2];
		
		for(; index<S.length && S[index]=='R'; index++)
			count[0]++;
		for(; index<S.length && S[index]=='L'; index++)
			count[1]++;
		
		return count;
	}
	
	static int[] getChild(int[] count) {
		int R = count[0];
		int L = count[1];
		int[] ans = new int[R+L];
		
		ans[R-1] = (R+1)/2 + L/2;
		ans[R] = R/2 + (L+1)/2;
		
		return ans;
	}
}