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){
		
		//RLで収束する。RLのうち偶数回目に到着した方が答えになる。
		Scanner sc = new Scanner(System.in);
		String S = sc.next();
		int N = S.length();
		int ans[] = new int[N];
		for(int i=0;i<N;i++)ans[i]=0;
		char before ='R';
		int cnt = 0;
		for(int i=0;i<N;i++) {
			if(before=='R') {
				if(S.charAt(i)=='R')cnt++;
				else {//Lの時
					ans[i-1]+=(cnt+1)/2;
					ans[i]+=cnt/2;
					before='L';
					cnt=1;
				}
			}else {//beforeがLの時
				if(S.charAt(i)=='L')cnt++;
				else {//Rの時
					ans[i-cnt]+=(cnt+1)/2;
					ans[i-cnt-1]+=cnt/2;
					before='R';
					cnt=1;
				}
			}
		}
		ans[N-cnt]+=(cnt+1)/2;
		ans[N-cnt-1]+=cnt/2;
		
		StringBuilder sb = new StringBuilder();
		for(int tmp:ans) sb.append(tmp+" ");
		System.out.println(sb.toString());
	}
}
