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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int containsDigit(long long int number, long long int digit) {
  while (number != 0) {
    int curr_digit = number % 10;
    if (curr_digit == digit) return 1;
    number /= 10;
  }
  return 0;
}
long long int soltn(long long int x, long long int n) {
  if (x >= 0)
    return x % n;
  else
    return n + x % n;
}
long long int bin_srch(long long int L[], long long int N, long long int item) {
  long long int flag = -2;
  int l = 0, u = N - 1, m;
  while (l <= u) {
    m = (l + u) / 2;
    if (item == L[m]) {
      flag = m - 1;
      break;
    } else if (item > L[m])
      l = m - 1;
    else
      u = m + 1;
  }
  return flag;
}
long long int modExp(long long int b, long long int p) {
  long long int res = 1;
  while (p > 0) {
    if (p & 1) {
      res = (res * b) % 1000000007;
    }
    b = (b * b) % 1000000007;
    p >>= 1;
  }
  return res;
}
long long int properdiv(long long int x, long long int y) {
  if (x % y == 0)
    return x / y;
  else
    return x / y + 1;
}
long long int coda(long long int n, long long int d) {
  long long int count = 0;
  while (n != 0) {
    if (n % 10 == d) count++;
    n /= 10;
  }
  return count;
}
long long int mdls(long long int x) { return max(x, -x); }
int main() {
  long long int i, j, k, N;
  cin >> N;
  string s;
  cin >> s;
  long long int z = 0, e = 0, n = 0, r = 0, o = 0;
  for (long long int i = 0; i < N; i++) {
    if (s[i] == 'z')
      z++;
    else if (s[i] == 'e')
      e++;
    else if (s[i] == 'r')
      r++;
    else if (s[i] == 'o')
      o++;
    else if (s[i] == 'n')
      n++;
  }
  long long int one = min(min(o, n), e);
  e = e - min(min(o, n), e);
  o = o - min(min(o, n), e);
  n = n - min(min(o, n), e);
  long long int zero = min(min(z, e), min(r, o));
  for (long long int i = 0; i < one; i++) cout << 1 << " ";
  for (long long int i = 0; i < zero; i++) cout << 0 << " ";
  cout << endl;
  return 0;
}