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) {
        // write your code here
        Scanner sc = new Scanner(System.in);

        String original = sc.next();
        int firstIndex = original.indexOf("RL");
        int nowIndex = 0;

        StringBuilder builder = new StringBuilder();

        while (firstIndex >= 0) {

            // 左辺の計算
            String left = original.substring(nowIndex, firstIndex);

            for (int i = 0; i < left.length(); i++) {
                builder.append("0 ");
            }

            int rCount = 1;
            int lCount = 1;
            rCount += left.length() / 2;
            lCount += Math.ceil((double) left.length() / 2);

            // 右辺の計算
            int nextIndex = firstIndex + 2 > original.length() ? - 1 : original.substring(firstIndex + 2).indexOf("RL");
            String right = "";
            // 右辺が存在する
            if (firstIndex + 3 <= original.length() && original.substring(firstIndex + 2, firstIndex + 3).equals("L")) {
                // まだRLがある
                if (nextIndex >= 0) {
                    nextIndex += firstIndex + 2;
                    right = original.substring(firstIndex + 2, nextIndex);
                    right = right.replace("R", "");
                    nowIndex = firstIndex + 2 + right.length();
                } else {
                    // もうない
                    right = original.substring(firstIndex + 2);
                    right = right.replace("R", "");
                    nowIndex = firstIndex + 2;
                }
                lCount += right.length() / 2;
                rCount += Math.ceil((double) right.length() / 2);
            } else {
                nowIndex = firstIndex + 2;
                if (nextIndex >= 0) {
                    nextIndex += firstIndex + 2;
                }
            }

            builder.append(rCount).append(" ").append(lCount).append(" ");

            for (int i = 0; i < right.length(); i++) {
                builder.append("0 ");
            }
            firstIndex = nextIndex;
        }
        System.out.println(builder);
    }
}