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
import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        char[] arr = sc.next().toCharArray();
        int n = arr.length;
        int lastR = 0;
        int[] rArr = new int[n];
        int lastL = n - 1;
        int[] lArr = new int[n];
        for (int i = 0; i < n; i++) {
            if (arr[i] == 'R') {
                lastR = i;
            }
            rArr[i] = lastR;
            if (arr[n - i - 1] == 'L') {
                lastL = n - i - 1;
            }
            lArr[n - i - 1] = lastL;
        }
        int[] ans = new int[n];
        for (int i = 0; i < n; i++) {
            if (arr[i] == 'R') {
                if ((lArr[i] - i) % 2 == 0) {
                    ans[lArr[i]]++;
                } else {
                    ans[lArr[i] - 1]++;
                }
            } else {
                if ((i - rArr[i]) % 2 == 0) {
                    ans[rArr[i]]++;
                } else {
                    ans[rArr[i] + 1]++;
                }
            }
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < n; i++) {
            if (i != 0) {
                sb.append(" ");
            }
            sb.append(ans[i]);
        }
        System.out.println(sb);
    }
}
