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<cstdio>
const int N=1e5;
int n,a[N+10],l,f[N+10],ans[N+10];
char s[N+10];
int main()
{
	scanf("%s",s);
	for(;s[n];n++)
		if(s[n]=='L')a[n+1]=-1;
		else a[n+1]=1;
	l=1;
	for(int i=1;i<n;i++)
		if(a[i]==1&&a[i+1]==-1)
		{
			for(int j=l;j<=i;j++)f[j]=i-j;
			for(int j=i+1;;j++)
			{
				if(j>n)
				{
					l=n;
					break;
				}
				if(a[j]!=-1)
				{
					l=j;
					break;
				}
				f[j]=i-j;
			}
		}
	for(int i=1,x;i<=n;i++)
		ans[i+f[i]+(N-f[i])%2]++;
	for(int i=1;i<=n;i++)
		printf("%d ",ans[i]);
	return 0;
}