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.*;

public class Cards {

    public static void main(String[] args) {
        MyScanner scn = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));

        // Start writing your solution here. -------------------------------------

        /*
         * int n = sc.nextInt(); // read input as integer long k = sc.nextLong(); //
         * read input as long double d = sc.nextDouble(); // read input as double String
         * str = sc.next(); // read input as String String s = sc.nextLine(); // read
         * whole line as String
         * 
         * int result = 3*n; out.println(result); // print via PrintWriter
         */

        // Stop writing your solution here. -------------------------------------
        int n = scn.nextInt();
        String str = scn.next();

        /*
         * int[] freq = new int[26];
         * 
         * for (int i = 0; i < str.length(); i++) { freq[str.charAt(i) - 'a']++; }
         * 
         * while (freq[13] != 0 && freq[4] != 0 && freq[14] != 0) { System.out.print(1 +
         * " "); freq[13]--; freq[4]--; freq[14]--;
         * 
         * } while (freq[25] != 0 && freq[4] != 0 && freq[17] != 0 && freq[14] != 0) {
         * System.out.print(0 + " "); freq[25]--; freq[4]--; freq[14]--; freq[17]--;
         * 
         * }
         */

        // 2nd better Method
        String ans = "";
        for (int i = 0; i < str.length(); i++) {
            if (str.charAt(i) == 'z')
                ans += "0 ";
            if (str.charAt(i) == 'n')
                ans = "1 " + ans;

        }
        System.out.println(ans);

        out.close();
    }

    public static PrintWriter out;

    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            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;
        }
    }

}