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;
const int MAXN=100010;
char c[MAXN];
int n,pre[MAXN],nx[MAXN],ans[MAXN]; 
int main () {
	scanf("%s",c+1);
	n=strlen(c+1);
	for (int i=1;i<=n;i++) {pre[i]=(c[i-1]=='R'?i-1:pre[i-1]);}
	for (int i=n;i>=1;i--) {nx[i]=(c[i+1]=='L'?i+1:nx[i+1]);}
	for (int i=1;i<=n;i++) {
		if (c[i]=='R') {
			if ((nx[i]-i)%2) {ans[nx[i]-1]++;}
			else {ans[nx[i]]++;}
		} else {
			if ((i-pre[i])%2) {ans[pre[i]+1]++;}
			else {ans[pre[i]]++;}
		}
	}
	for (int i=1;i<=n;i++) {
		printf("%d ",ans[i]);
	}
	return 0;
}