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
#include <bits/stdc++.h>
using namespace std;
string s;
int k[100010];
int main()
{
	cin>>s;
	for(int i=0;i<s.length()-1;i++)
	{
		int l=0,r=0;
		int pos=0;
		if (s[i]=='R'&&s[i+1]=='L')
		{
			for (int j=i-1;j>=0;j--)
			{
				if (s[j]=='R') l ++;
				else break;
			}
			for (pos=i+2;pos<s.length();pos ++)
			{
				if (s[pos]=='L') r ++;
				else break;
			}
			k[i]=l/2+r-r/2+1;
			k[i+1]=l-l/2+r/2+1;
			i=pos-1;
		}
	}
	for (int i=0;i<s.length();i++) printf("%d ",k[i]);
	printf("\n");
}
