description: You are given an integer n. Find any string s of length n consisting only of English lowercase letters such that each non-empty substring of s occurs in s an odd number of times. If there are multiple such strings, output any. It can be shown that such string always exists under the given constraints.
  
  A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 500) — the number of test cases.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
  
  It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
  
  Output
  
  For each test case, print a single line containing the string s. If there are multiple such strings, output any. It can be shown that such string always exists under the given constraints.
  
  Example
  
  Input
  
  
  4
  3
  5
  9
  19
  
  
  Output
  
  
  abc
  diane
  bbcaabbba
  youarethecutestuwuu
  
  Note
  
  In the first test case, each substring of "abc" occurs exactly once.
  
  In the third test case, each substring of "bbcaabbba" occurs an odd number of times. In particular, "b" occurs 5 times, "a" and "bb" occur 3 times each, and each of the remaining substrings occurs exactly once.
solution: 
  import java.io.*;
  import java.lang.reflect.Field;
  import java.util.*;
  
  
  public class test {
  
      public static void main(String[] args) {
          FastScanner sc = new FastScanner();
          int t = sc.ni();
          while (t-- > 0) {
              int n = sc.ni();
              StringBuilder sb = new StringBuilder();
              for (int i = 0; i < n / 2; i++)
                  sb.append('a');
              sb.append('b');
              for (int i = 0; i < n / 2 - 1; i++)
                  sb.append('a');
              if (n > 1 && n % 2 == 1) {
                  sb.append('c');
              }
              System.out.println(sb.toString());
          }
      }
  
      //
  
      static class FastScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public FastScanner() {
              br = new BufferedReader(new InputStreamReader(System.in), 32768);
              st = null;
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int ni() {
              return Integer.parseInt(next());
          }
  
          int[][] graph(int N, int[][] edges) {
              int[][] graph = new int[N][];
              int[] sz = new int[N];
              for (int[] e : edges) {
                  sz[e[0]] += 1;
                  sz[e[1]] += 1;
              }
              for (int i = 0; i < N; i++) {
                  graph[i] = new int[sz[i]];
              }
              int[] cur = new int[N];
              for (int[] e : edges) {
                  graph[e[0]][cur[e[0]]] = e[1];
                  graph[e[1]][cur[e[1]]] = e[0];
                  cur[e[0]] += 1;
                  cur[e[1]] += 1;
              }
              return graph;
          }
  
          int[] intArray(int N, int mod) {
              int[] ret = new int[N];
              for (int i = 0; i < N; i++)
                  ret[i] = ni() + mod;
              return ret;
          }
  
          long nl() {
              return Long.parseLong(next());
          }
  
          long[] longArray(int N, long mod) {
              long[] ret = new long[N];
              for (int i = 0; i < N; i++)
                  ret[i] = nl() + mod;
              return ret;
          }
  
          double nd() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
  }
  