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 = 1e5 + 5;
int a[maxn];
char s[maxn];

int main() {

    scanf("%s", s);
    int R = 0, L = 0, Ln, Rn;
    while(s[R]) {

        while(s[R] == 'R') R++;
        Ln = R - L; L = R+1;
        while(s[L] == 'L') L++;
        Rn = L - R;

        a[R] = a[R-1] = (Ln+Rn)/2;
        if((Ln+Rn)&1)
            if(Ln > Rn && Ln&1 || Ln < Rn && !(Rn&1)) a[R-1]++;
            else a[R]++;

        R = L;
    }
    for(int i = 0; s[i]; i++)
        printf("%d ", a[i]);
    return 0;
}