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;
typedef long long ll;





int main(){
	string S;
	cin >> S;
	int len = S.length();
	vector<int> v(len);
	
	S = S + "X";
	
	for ( int i = 0; i < len; ){
		int i0 = i;
		
		int R = 0, L = 0;
		// RR...RLL...L の連続数を調べる
		while ( S[i] == 'R' ){
			R++;
			i++;
		}
		while ( S[i] == 'L' ){
			L++;
			i++;
		}
		
		// RLの境界に集まる
		v[i0+R-1]   = (R+1)/2 + L/2;
		v[i0+R] = R/2 + (L+1)/2;
	}
	
	for ( auto it : v )
		cout << it << " ";
	cout << endl;
	
	return 0;
}

