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;
char S[100005];
int ANS[100005],n;
void DO(int s,int e) {
	for(int i=e+1; i<=n&&S[i]=='L'; i++) {
		if((i-e)&1)ANS[s]++;
		else ANS[e]++;
	}
	for(int i=s-1; i&&S[i]=='R'; i--) {
		if((s-i)&1)ANS[e]++;
		else ANS[s]++;
	}
}
int main() {
	scanf("%s",S+1);
	n=strlen(S+1);
	for(int i=1; i<=n; i++)
		if(S[i]=='R'&&S[i+1]=='L') DO(i,i+1),ANS[i]++,ANS[i+1]++;
	for(int i=1; i<=n; i++)printf("%d ",ANS[i]);
	return 0;
}