Problem: When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one. 

Input

The first line contains a single integer n (1 ⩽ n ⩽ 10^5) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.

It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 0 or "one" which corresponds to the digit 1.

Output

Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.

Examples

Input


4
ezor


Output


0 


Input


10
nznooeeoer


Output


1 1 0 

Note

In the first example, the correct initial ordering is "zero".

In the second example, the correct initial ordering is "oneonezero".

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.text.*;
import java.math.*;
import static java.lang.Integer.*;
import static java.lang.Double.*;
import java.lang.Math.*;

public class cards {

	public static void main(String[] args) throws Exception {
		new cards().run();
	}

	public void run() throws Exception {
		FastIO file = new FastIO();
		int z = 0;
		int n = file.nextInt();
		char[] c = file.next().toCharArray();
		for (int i = 0; i < n; i++) {
			if (c[i] == 'z') z++;
		}
		int zero = z * 4;
		int one = n - (zero);
		int o = one / 3;
		for (int i = 0; i < o; i++) System.out.print("1" + " ");
		for (int j = 0; j < z; j++) System.out.print("0" + " ");
	}

	public static class FastIO {
		BufferedReader br;
		StringTokenizer st;

		public FastIO() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	public static long pow(long n, long p, long mod) {
		if (p == 0)
			return 1;
		if (p == 1)
			return n % mod;
		if (p % 2 == 0) {
			long temp = pow(n, p / 2, mod);
			return (temp * temp) % mod;
		} else {
			long temp = pow(n, (p - 1) / 2, mod);
			temp = (temp * temp) % mod;
			return (temp * n) % mod;

		}
	}

	public static long pow(long n, long p) {
		if (p == 0)
			return 1;
		if (p == 1)
			return n;
		if (p % 2 == 0) {
			long temp = pow(n, p / 2);
			return (temp * temp);
		} else {
			long temp = pow(n, (p - 1) / 2);
			temp = (temp * temp);
			return (temp * n);

		}
	}

	public static long gcd(long x, long y) {
		if (x == 0)
			return y;
		else
			return gcd(y % x, x);
	}

	public static boolean isPrime(int n) {
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;

		if (n % 2 == 0 || n % 3 == 0)
			return false;

		for (int i = 5; i * i <= n; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;

		return true;
	}
}