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.*;
import java.math.*;
import java.lang.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		// 整数の入力
		String str = sc.next();
		String[] strArray = str.split("");

		boolean rightTurn = true;
		int leftCount = 0;
		int rightCount = 0;
		// System.out.println("strArray.length " + strArray.length);
		for (int i = 0; i < strArray.length; i++) {
			String s = strArray[i];
			boolean isRight = s.equals("R");
			// System.out.println("s " + s);

			if(rightTurn && isRight){
				rightCount ++;
			} else if (!rightTurn && !isRight){
				leftCount ++;
			} else if (rightTurn && !isRight) {
				leftCount++;
				rightTurn = false;
			} else {
				rightTurn = true;
				// ここが区切り

				if((leftCount + rightCount)%2 == 0 ){
					int sum = leftCount + rightCount;
					for (int j = 0; j < rightCount-1; j++) {
						System.out.print("0 ");
					}
					System.out.print((sum/2) + " ");
					System.out.print((sum/2) + " ");
					for (int j = 0; j < leftCount-1; j++) {
						System.out.print("0 ");
					}
				} else {
					// 奇数は面倒
					int sum = leftCount + rightCount;
					for (int j = 0; j < rightCount-1; j++) {
						System.out.print("0 ");
					}
					if(rightCount % 2 == 0){
						int result = sum/2;
						System.out.print(result + " ");
					}else{
						int result = sum/2 +1;
						System.out.print(result + " ");
					}


					if(leftCount % 2 == 0){
						int result = sum/2;
						System.out.print(result + " ");
					}else{
						int result = sum/2 +1;
						System.out.print(result + " ");
					}
					for (int j = 0; j < leftCount-1; j++) {
						System.out.print("0 ");
					}
				}
				leftCount = 0;
				rightCount = 1;
			}
		}


		// System.out.println();
		// System.out.println("rightCount " + rightCount);
		// System.out.println("leftCount " + leftCount);
		if((leftCount + rightCount)%2 == 0 ){
			int sum = leftCount + rightCount;
			for (int j = 0; j < rightCount-1; j++) {
				System.out.print("0 ");
			}
			System.out.print((sum/2) + " ");
			System.out.print((sum/2) + " ");
			for (int j = 0; j < leftCount-1; j++) {
				System.out.print("0 ");
			}
		} else {
			// 奇数は面倒
			int sum = leftCount + rightCount;
			for (int j = 0; j < rightCount-1; j++) {
				System.out.print("0 ");
			}
			if(rightCount % 2 == 0){
				int result = sum/2;
				System.out.print(result + " ");
			}else{
				int result = sum/2 +1;
				System.out.print(result + " ");
			}


			if(leftCount % 2 == 0){
				int result = sum/2;
				System.out.print(result + " ");
			}else{
				int result = sum/2 +1;
				System.out.print(result + " ");
			}
			for (int j = 0; j < leftCount-1; j++) {
				System.out.print("0 ");
			}
		}

	}

}
