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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
 
public class Solution2 {
  public static void main(String args[]){
    InputReader2 inp = new InputReader2(System.in);
    solve(inp);
  }
  
  public static void solve(InputReader2 in) {
    int num = in.nextInt();
    int n = 0, z = 0;
    String str = in.next();
    for (int i = 0; i < num; i++) {
      if (str.charAt(i) == 'n') {
        n++;
      } else if (str.charAt(i) == 'z') {
        z++;
      }
    }
    for (int i = 0; i < n; i++) {
      System.out.print("1 ");
    }
    for (int i = 0; i < z; i++) {
      System.out.print("0 ");
    }    
  }
}
 
class InputReader2 {
  
  public BufferedReader bufferedReader;
  public StringTokenizer stringTokenizer;
  
  public InputReader2(InputStream in) {
    bufferedReader = new BufferedReader(new InputStreamReader(in));
    stringTokenizer = null;
  }
  
  public String next() {
    String str;
    if(stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
      try {
        stringTokenizer = new StringTokenizer(bufferedReader.readLine());    
      } catch (IOException e) {
      System.out.print(e.getStackTrace());
      }
    }
    return stringTokenizer.nextToken();
  }
  
  public int nextInt() {
    return Integer.parseInt(this.next());
  }
 
  public long nextLong() {
    return Long.parseLong(this.next());
  }
  
}







/*    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    String command = sc.next();
    int l = 0,r = 0;
    for (int i = 0; i < command.length(); i++) {
      if (command.charAt(i) == 'L') {
        l++;
      } else {
        r++;
      }
    }
    System.out.print(l+r+1);
*/