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;
#define maxn 100005
char s[maxn];
int f[maxn][2];
int main(){
	scanf("%s",s+1);
	int i;
	for(i=1;s[i];i++){
		f[i][0]=1;
		if(s[i-1]=='R'&&s[i]=='R'){
			f[i][0]+=f[i-1][1],f[i][1]+=f[i-1][0];
		}
	}
	for(i=i-1;s[i];i--){
		if(s[i+1]=='L'&&s[i]=='L'){
			f[i][0]+=f[i+1][1],f[i][1]+=f[i+1][0];
		}
	}
	for(int i=1;s[i];i++){
		if(s[i]=='R'&&s[i+1]=='L'){
			printf("%d %d ",f[i][0]+f[i+1][1],f[i][1]+f[i+1][0]);i++;
		}
		else printf("0 ");
	}
	return 0;
}