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
//解説放送を見て書き直す
#include<bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(n);++i)
using namespace std;
typedef long long ll;

int main(){
	string s;
	cin >> s;
	int n;
	n = (int)s.length();
	vector<int> ans(n);
	rep(ri,2){
		{	
			int cnt=0;
			rep(i,n){
				if(s[i] == 'R') cnt++;
				else{
					ans[i] += cnt/2;
					ans[i-1] += (cnt+1)/2;
					cnt = 0;
				}
			}
		}
		reverse(ans.begin(),ans.end());
		reverse(s.begin(),s.end());
		rep(i,n){
			if(s[i] == 'L') s[i] = 'R';
			else s[i] = 'L';
		}
	}
	rep(i,n){
		printf("%d\n",ans[i]);
	}
	return 0;
}
