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 Solution{
    public static class pair{
        int x;
        int y;
        int z;
    }
    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(br.readLine());
        String line=br.readLine();
        // String[] strs=line.trim().split(" ");
        // int n=Integer.parseInt(strs[0]);
        // int L=Integer.parseInt(strs[1]);
        // int a=Integer.parseInt(strs[2]);
        // int[][] arr=new int[n][2];
        // if(n==0){
        //     System.out.println(L/a);
        //     return;
        // }
        int one=0;
        int zero=0;
        for(int i=0;i<n;i++){
            // line=b[1]=Integer.parseInt(strs[1]);
            if(line.charAt(i)=='n'){
                one++;
            }else if(line.charAt(i)=='z'){
                zero++;
            }
        }
        for(int i=0;i<one;i++){
            System.out.print("1 ");
        }
        
        for(int i=0;i<zero;i++){
            System.out.print("0 ");
        }
        System.out.println();
        // long c=0;
        // for(int i=0;i<arr.length-1;i++){
        //     long val=arr[i+1][0]-(arr[i][0]+arr[i][1]);
        //     c+=(val/a);
        // }
        // long ready=L-(arr[arr.length-1][0]+arr[arr.length-1][1]);
        // c+=(ready)/a;
        // System.out.println(c);
        
    }
    public static boolean prime(long val){
        for(long i=2;i<val;i++){
            if(val%i==0){
                return false;
            }
        }
        return true;
    }
    public static class Comp implements Comparator<pair>{
        @Override
        public int compare(pair a,pair b){
            if(a.y==b.y){
                return b.x-a.x;
            }else{
                return a.y-b.y;
            }
        }
    }
    public static class Com implements Comparator<pair>{
        @Override
        public int compare(pair a,pair b){
            if(a.x!=b.x){
                return a.x-b.x; 
            }else{
                return a.y-b.y;
            }
        }
    }
    public static void pr(int v1,int v2){
        int v3=v1;
        int v4=v2;
        // System.out.println(v1+" "+v2);
        for(int i=2;i<=Math.min(v3,v4);i++){
            while(v1%i==0&&v2%i==0){
                v1/=i;
                v2/=i;
            }
        }
        System.out.println(v1+"/"+v2);
    }
    public static int lcm(int a,int b){
        int x=Math.max(a,b);
        int y=Math.min(a,b);
        long ans=x;
        while(ans%y!=0){
            ans+=x;
        }
        return (int)ans;
    }
    public static long fact(int n){
        long ans=1;
        for(int i=1;i<=n;i++){
            ans*=i;
        }
        return ans;
    }
}