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.*;
import static java.lang.Math.*;

public class Main implements Runnable
{
    boolean multiple = false;
    long MOD = 998244353;

    @SuppressWarnings("Duplicates")
    void solve() throws Exception
    {
        int n = sc.nextInt();
        String s = sc.nextToken();
        int one = 0, zero = 0;
        for (int i = 0; i < n; i++)
            if (s.charAt(i) == 'n') one++; else if (s.charAt(i) == 'z') zero++;
        for (int i = 0; i < one; i++)
            p(1 + " ");
        for (int i = 0; i < zero; i++)
            p(0 + " ");
        pl();
    }

    StringBuilder ANS = new StringBuilder();
    void p(String s) { ANS.append(s); } void p(double s) {ANS.append(s); } void p(long s) {ANS.append(s); } void p(char s) {ANS.append(s); }
    void pl(String s) { ANS.append(s); ANS.append('\n'); } void pl(double s) { ANS.append(s); ANS.append('\n'); } void pl(long s) { ANS.append(s); ANS.append('\n'); } void pl(char s) { ANS.append(s); ANS.append('\n'); } void pl() { ANS.append(('\n')); }
    /*I/O, and other boilerplate*/ @Override public void run() { try { in = new BufferedReader(new InputStreamReader(System.in));out = new PrintWriter(System.out);sc = new FastScanner(in);if (multiple) { int q = sc.nextInt();for (int i = 0; i < q; i++) solve(); } else solve(); System.out.print(ANS); } catch (Throwable uncaught) { Main.uncaught = uncaught; } finally { out.close(); }} public static void main(String[] args) throws Throwable{ Thread thread = new Thread(null, new Main(), "", (1 << 26));thread.start();thread.join();if (Main.uncaught != null) {throw Main.uncaught;} } static Throwable uncaught; BufferedReader in; FastScanner sc; PrintWriter out; } class FastScanner { BufferedReader in; StringTokenizer st; public FastScanner(BufferedReader in) {this.in = in;}public String nextToken() throws Exception { while (st == null || !st.hasMoreTokens()) { st = new StringTokenizer(in.readLine()); }return st.nextToken(); }public int nextInt() throws Exception { return Integer.parseInt(nextToken()); }public long nextLong() throws Exception { return Long.parseLong(nextToken()); }public double nextDouble() throws Exception { return Double.parseDouble(nextToken()); }
}