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 <cstring>
using namespace std;

const int N = 100 * 1000 + 5;
int n, a[N], L[N], R[N];
char s[N];

int main()
{
    scanf("%s",s+1); n = strlen(s+1);
    for(int i=n; i>=1; --i) R[i] = (s[i] == 'R') ? R[i+1] : i;
    for(int i=1; i<=n; ++i) L[i] = (s[i] == 'L') ? L[i-1] : i;
    for(int i=1; i<=n; ++i) if (s[i] == 'R') {
        if ((R[i]-i) & 1) a[R[i]-1] += 1; else a[R[i]] += 1;
    } else {
        if ((i-L[i]) & 1) a[L[i]+1] += 1; else a[L[i]] += 1;
    }
    for(int i=1; i<=n; ++i) printf("%d ",a[i]);
}
