Problem: IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting.

Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from 1 to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing?

A string a is lexicographically smaller than a string b if and only if one of the following holds:

  * a is a prefix of b, but a ≠ b;
  * in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.

Input

The first and the only line contains a string s (1 ≤ |s| ≤ 1000), describing the initial string formed by magnets. The string s consists only of characters 'a' and 'b'.

Output

Output exactly |s| integers. If IA should reverse the i-th prefix (that is, the substring from 1 to i), the i-th integer should be equal to 1, and it should be equal to 0 otherwise.

If there are multiple possible sequences leading to the optimal answer, print any of them.

Examples

Input

bbab


Output

0 1 1 0


Input

aaaaa


Output

1 0 0 0 1

Note

In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string.

In the second example, she can reverse any subset of prefixes — all letters are 'a'.

Solution (in JAVA):

Solution:

import java.io.*;

import java.util.*;
import java.util.Map.Entry;

import javax.swing.Box.Filler;
import javax.swing.text.InternationalFormatter;

import java.math.*;

public class temp1 {

	public static PrintWriter out;
	static int o = 0, e = 0;
	static int[] fa;
	static int[][][][] dp;
	static ArrayList<pair> p = new ArrayList<>();
	static int ans = 0;

	public static void main(String[] args) throws IOException {

		// Reader scn = new Reader();
		Scanner scn = new Scanner(System.in);

		StringBuilder sb = new StringBuilder();

		String s = scn.next();
		boolean b = false;

		for (int i = 0; i < s.length(); i++) {

			if (i == s.length() - 1 && s.charAt(i) == 'a')
				sb.append("1");

			else if (i + 1 < s.length() && s.charAt(i) == 'b' && s.charAt(i + 1) == 'a') {
				sb.append("1 ");
				b = true;
			}

			else if (i + 1 < s.length() && s.charAt(i) == 'a' && s.charAt(i + 1) == 'b' && b) {
				sb.append("1 ");
				b = false;
			}

			else
				sb.append("0 ");
		}

		System.out.println(sb);

	}

	// _________________________TEMPLATE_____________________________________________________________

	// private static int gcd(int a, int b) {
	// if(a== 0)
	// return b;
	//
	// return gcd(b%a,a);
	// }

	// static class comp implements Comparator<pair> {
	//
	// @Override
	// public int compare(pair o1, pair o2) {
	//
	// return (int) (o1.y - o1.y);
	// }
	//
	// }

	private static void fillit(int[] a, int i) {

		int idx = i - 1;

		while (idx >= 0 && a[idx] == 0 && o > 0) {
			a[idx--] = 1;
			o--;
		}

		idx = i + 1;

		while (idx < a.length && a[idx] == 0 && o > 0) {
			a[idx++] = 1;
			o--;
		}
	}

	private static void Fill(int[] a, int x, int y) {

		for (int i = x; i <= y; i++)
			a[i] = 1;
	}

	static class pair implements Comparable<pair> {
		int x = 0;
		int y = 0;
		int d = 0;

		pair(int a, int b) {
			x = a;
			y = b;
			d = y - x;
		}

		@Override
		public int compareTo(pair o) {

			return this.d - o.d;
		}
	}

	public static class Reader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;

		public Reader() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public Reader(String file_name) throws IOException {
			din = new DataInputStream(new FileInputStream(file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public String readLine() throws IOException {
			byte[] buf = new byte[100000 + 1]; // line length
			int cnt = 0, c;
			while ((c = read()) != -1) {
				if (c == '\n')
					break;
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, cnt);
		}

		public int nextInt() throws IOException {
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}

		public long nextLong() throws IOException {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}

		public double nextDouble() throws IOException {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (c == '.') {
				while ((c = read()) >= '0' && c <= '9') {
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}

		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}

		private byte read() throws IOException {
			if (bufferPointer == bytesRead)
				fillBuffer();
			return buffer[bufferPointer++];
		}

		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}

		public int[] nextIntArray(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nextInt();
			}
			return arr;
		}

		public long[] nextLongArray(int n) throws IOException {
			long[] arr = new long[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nextLong();
			}
			return arr;
		}

		public int[][] nextInt2DArray(int m, int n) throws IOException {
			int[][] arr = new int[m][n];
			for (int i = 0; i < m; i++) {
				for (int j = 0; j < n; j++)
					arr[i][j] = nextInt();
			}
			return arr;
		}

	}

}