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 <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;

int main(){
	long i=0,n,l,s,r;
	string str;
	cin>>str;
	n=str.size();
	vector<long> chi(n,0);
	while(i<n){
		l=i;
		while(str.at(i)=='R'){
			i++;
		}
		s=i;
		while(str.at(i)=='L'){
			i++;
			if(i==n){
				break;
			}
		}
		r=i-1;
		//cout<<l<<s<<r<<endl;
		chi.at(s-1)+=(s-l)/2+(s-l)%2;
		chi.at(s)+=(s-l)/2;
		chi.at(s-1)+=(r-s+1)/2;
		chi.at(s)+=(r-s+1)/2+(r-s+1)%2;
	}
	cout<<chi.at(0);
	for(i=1;i<n;i++){
		cout<<" "<<chi.at(i);
	}
	return 0;
}

