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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
#define MAXN 100000
int n,p,cntL,cntR,pos;
char s[MAXN+5];
int ans[MAXN+5];
int main()
{
	scanf("%s",s+1);
	n=strlen(s+1);
	s[n+1]='R';
	for(int i=1;i<=n+1;i++)
	if(s[i]=='R'){
		if(!p){
			p=1;
			ans[pos-1]+=(cntL+1)/2+cntR/2;
			ans[pos]+=(cntR+1)/2+cntL/2;
			cntL=cntR=0;
		}
		cntL++;
	}else{
		if(p)pos=i;
		p=0;
		cntR++;
	}
	for(int i=1;i<=n;i++)
	printf("%d ",ans[i]);
}