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.*;

public class Cards{
  public static void main(String[] args){
    Scanner gd = new Scanner(System.in);
    int tt = 1;
    while(tt>0){
      int num = IO.ri();
      String str = IO.rs();
      long z=0,e=0,r=0,o=0,n=0;
      long ones = 0,zeros = 0;
      for(int i=0;i<num;i++){
        if(Character.toString(str.charAt(i)).equals("n")){
          n++;
        }
        else if(Character.toString(str.charAt(i)).equals("o")){
          o++;
        }
        else if(Character.toString(str.charAt(i)).equals("r")){
          r++;
        }
        else if(Character.toString(str.charAt(i)).equals("e")){
          e++;
        }
        else if(Character.toString(str.charAt(i)).equals("z")){
          z++;
        }
      }
      Long min = (long)Math.min(n,e);
      min = (long)Math.min(min,o);
      ones = min;
      e-=ones;
      o-=ones;
      min = (long)Math.min(e,o);
      min = (long)Math.min(min,r);
      min = (long)Math.min(min,z);
      zeros = min;
      for(int g=0;g<ones;g++){
        IO.pf("1 ");
      }
      for(int g=0;g<zeros;g++){
        IO.pf("0 ");
      }
      IO.pf("\n");
      tt--;
    }
  }
}

// Template
class IO{
  static Scanner gd = new Scanner(System.in);
  public static long rl(){
    return gd.nextLong();
  }
  public static int ri(){
    return gd.nextInt();
  }
  public static String rs(){
    return gd.next();
  }
  public static int[] rai(int len){
    int[] temp = new int[len];
    for(int i=0;i<len;i++){
      temp[i] = gd.nextInt();
    }
    return temp;
  }
  public static long[] ral(int len){
    long[] temp = new long[len];
    for(int i=0;i<len;i++){
      temp[i] = gd.nextLong();
    }
    return temp;
  }
  public static String[] ras(int len){
    String[] temp = new String[len];
    for(int i=0;i<len;i++){
      temp[i] = gd.next();
    }
    return temp;
  }
  public static void pf(String format,Object... args){
    System.out.printf(format,args);
  }
  public static void pl(Object data){
    System.out.println(data);
  }
  public static void pai(int arr[]){
    for(int i=0;i<arr.length;i++){
      System.out.printf("%d ",arr[i]);
    }
    System.out.println();
  }
  public static void pal(long arr[]){
    for(int i=0;i<arr.length;i++){
      System.out.printf("%d ",arr[i]);
    }
    System.out.println();
  }
  public static void pas(String arr[]){
    for(int i=0;i<arr.length;i++){
      System.out.printf("%s ",arr[i]);
    }
    System.out.println();
  }
}