In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick.

He is too selfish, so for a given n he wants to obtain a string of n characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible.

Input

The first line contains single integer n (1 ≤ n ≤ 2·105) — the length of the string.

Output

Print the string that satisfies all the constraints.

If there are multiple answers, print any of them.

Examples

Input

2


Output

aa


Input

3


Output

bba

Note

A palindrome is a sequence of characters which reads the same backward and forward.
#include <bits/stdc++.h>
using namespace std;
long long int visited[51];
long long int Fact(long long int x) {
  long long int y = 1;
  for (long long int i = 1; i <= x; i++) {
    y = ((y % 998244353) * (i % 998244353)) % 998244353;
  }
  return y;
}
long long int dfs(long long int parent, vector<long long int> v[]) {
  long long int y = 1;
  visited[parent] = 1;
  for (long long int i = 0; i < v[parent].size(); i++) {
    if (visited[v[parent][i]] == 0) {
      y += dfs(v[parent][i], v);
    }
  }
  return y;
}
long long int modularExponentiation(long long int x, long long int n,
                                    long long int M) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = ((result % M) * (x % M)) % M;
    x = ((x % M) * (x % M)) % M;
    n = n / 2;
  }
  return result;
}
long long int binaryExponentiation(long long int x, long long int n) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = result * x;
    x = x * x;
    n = n / 2;
  }
  return result;
}
long long int pow1(long long int x, long long int y) {
  long long int z = 1;
  while (y--) {
    z = z * x;
  }
  return z;
}
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long int check(long long int x) {
  long long int y = 1, z = 0;
  while (x > y) {
    y *= 2;
    z++;
  }
  if (x != y) {
    z--;
  }
  return z;
}
unsigned long long modInverse(unsigned long long n, int p) {
  return modularExponentiation(n, p - 2, p);
}
unsigned long long nCrModPFermat(unsigned long long n, int r, int p) {
  if (r == 0) return 1;
  unsigned long long fac[n + 1];
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
long long int modInverse(long long int A, long long int M) {
  return modularExponentiation(A, M - 2, M);
}
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  if (a.first == b.first)
    return a.second < b.second;
  else
    return a.first < b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    if ((i % 4) < 2)
      cout << 'a';
    else
      cout << 'b';
  }
}
