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<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
const ll M=998244353;
char c[100005];
int ans[100005],i,j,s[2];
int main(){
	scanf("%s",c+1);
	for(i=1;c[i];i++)
		if(c[i]=='R'&&c[i+1]=='L')
		{
			memset(s,0,sizeof(s));
			for(j=i;c[j]=='R';j--)
				s[(i-j)%2]++;
			for(j=i+1;c[j]=='L';j++)
				s[(j-i)%2]++;
			ans[i]=s[0];
			ans[i+1]=s[1];
		}
	for(i=1;c[i];i++)
		printf("%d ",ans[i]);
}
