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.io.*;
import java.util.*;

public class A {

	void run() {
		int n = in.nextI();
		String s = in.next();
		int i;
		int c0 = 0, c1 = 0;
		for(i=0;i<n;i++){
			if(s.charAt(i) == 'z') c0++;
			if(s.charAt(i) == 'n') c1++;
		}

		for(i=0;i<c1;i++)
			out.print("1 ");

		for(i=0;i<c0;i++)
			out.print("0 ");

	}
	
	static MyScanner in;
	static PrintWriter out;

	public static void main(String[] args) throws IOException {
		boolean stdStream = true;
		if(stdStream){
			in = new MyScanner();
			out = new PrintWriter(System.out);
		}else{
			String fileName = A.class.getSimpleName();
			int n_test = 1;			
			String inputFileName = fileName + String.format(".%02d", n_test) + ".inp";
			String outputFileName = fileName + String.format(".%02d", n_test) + ".out";
			in = new MyScanner(new BufferedReader(new FileReader(inputFileName)));
			out = new PrintWriter(outputFileName);
		}
			
		new A().run();

		in.close();
		out.close();
	}

	static class MyScanner {
		BufferedReader br;
		StringTokenizer st;

        MyScanner(){
            this.br = new BufferedReader(new InputStreamReader(System.in));
        }

		MyScanner(BufferedReader br) {
			this.br = br;
        }

        void close() throws IOException {
            br.close();
        }
        
		void findToken() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
		}
		
		String next() {
			findToken();
			return st.nextToken();
		}
		
		int nextI() {
			return Integer.parseInt(next());
		}

		int[] nextI(int n) {
			int v[] = new int[n];
			for(int i=0;i<n;i++) v[i] = nextI();
			return v;
		}
		
		long nextL() {
			return Long.parseLong(next());
		}

		long[] nextL(int n) {
			long v[] = new long[n];
			for(int i=0;i<n;i++) v[i] = nextL();
			return v;
		}
		
		double nextD() {
			return Double.parseDouble(next());
		}

		double[] nextD(int n) {
			double v[] = new double[n];
			for(int i=0;i<n;i++) v[i] = nextD();
			return v;
		}
	}

	class Pair{
		int a,b;
		Pair(int a, int b){ this.a = a; this.b = b;}
	}
}