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>
#include<cstring>

#define ran 101101

int n;
char s[ran];
int to[ran], ans[ran];

int main() {
	scanf("%s", s);
	n = strlen(s);
	
	memset(to, -1, sizeof to);
	for(int i=0;i<n-1;i++) {
		if (s[i]!='R' || s[i+1]!='L') continue;
		to[i] = to[i+1] = i;
		
		for(int j=i-1;j>=0 && s[j] == 'R'; j--)
			to[j] = i;
		for(int j=i+2;j<n && s[j] == 'L'; j++)
			to[j] = i;
	}
	
	for(int i=0;i<n;i++)
		if ((to[i]-i)%2) ans[to[i]+1] ++;
		else ans[to[i]]++;
	for(int i=0;i<n;i++)
		printf("%d%c", ans[i], i<n-1 ? ' ' : '\n');
	
	return 0;
}