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;

int main(){
	int i;
	string S;
	cin >> S;
	vector<int> ans(S.size(), 0);
	for(i = 0; i < S.size(); ){
		int j, k;
		for(j = i; j < S.size() && S[j] == 'R'; j++);
		for(k = j; k < S.size() && S[k] == 'L'; k++);
//		printf("(i, j, k) = (%d, %d, %d)\n", i, j, k);
		ans[j - 1] = k / 2 - i / 2;
		ans[j] = k - i - ans[j - 1];
		if(j % 2 == 1){
			swap(ans[j - 1], ans[j]);
		}
		i = k;
	}
	for(i = 0; i < S.size(); i++){
		printf("%d ", ans[i]);
	}
	printf("\n");
	return 0;
}