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.*;
import java.util.stream.*;

class Main{
	static int mod=1000000007;

	public static void main(String[] $){
		System.err.println();
		Scanner s=new Scanner(System.in);

		char[]t=s.next().toCharArray();
		int n=t.length;
		int[]a=new int[n];
		int[]b=new int[n];
		Arrays.fill(a,1);
		Arrays.fill(b,0);
		for(int i=0;i<n;++i) {
			if(t[i]=='R'&&t[i+1]=='R') {
				a[i+1]+=b[i];
				b[i+1]+=a[i];
				a[i]=0;
				b[i]=0;
			}
		}
		for(int i=n-1;i>=0;--i) {
			if(t[i]=='L'&&t[i-1]=='L') {
				a[i-1]+=b[i];
				b[i-1]+=a[i];
				a[i]=0;
				b[i]=0;
			}
		}

		for(int i=0;i<n;++i) {
			if(t[i]=='R'&&t[i+1]!='R') {
				a[i+1]+=b[i];
			}
		}
		for(int i=n-1;i>=0;--i) {
			if(t[i]=='L'&&t[i-1]!='L') {
				a[i-1]+=b[i];
			}
		}

		System.out.println(Arrays.stream(a).mapToObj(String::valueOf).collect(Collectors.joining(" ")));
	}
}
