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: #include <bits/stdc++.h>
  using namespace std;
  struct hash_pair {
    template <class T1, class T2>
    size_t operator()(const pair<T1, T2> &p) const {
      auto hash1 = hash<T1>{}(p.first);
      auto hash2 = hash<T2>{}(p.second);
      return hash1 ^ hash2;
    }
  };
  long long int n, m;
  vector<long long int> adj[1];
  bool visited[1];
  long long int start[1];
  long long int end[1];
  long long int parent[1];
  long long int level[1];
  vector<long long int> dfs_order;
  void dfs(long long int node) {
    visited[node] = true;
    for (long long int next : adj[node]) {
      if (!visited[next]) {
        parent[next] = node;
        level[next] = level[node] + 1;
        dfs(next);
      }
    }
  }
  long long int dist[1];
  void bfs(long long int start) {
    memset(dist, -1, sizeof dist);
    queue<long long int> q;
    dist[start] = 0;
    q.push(start);
    while (!q.empty()) {
      long long int v = q.front();
      q.pop();
      for (long long int e : adj[v]) {
        if (dist[e] == -1) {
          dist[e] = dist[v] + 1;
          q.push(e);
        }
      }
    }
  }
  long long int lift(long long int a, long long int dist,
                     vector<vector<long long int>> &up) {
    for (long long int i = 0; i < 20; i++) {
      if (dist & (1 << i)) {
        a = up[a][i];
      }
    }
    return a;
  }
  void preprocesslift(vector<vector<long long int>> &up) {
    for (long long int j = 1; j < 20; j++) {
      for (long long int i = 1; i <= n; i++) {
        if (up[i][j - 1] != -1) {
          up[i][j] = up[up[i][j - 1]][j - 1];
        }
      }
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t;
    cin >> t;
    while (t > 0) {
      long long int n;
      cin >> n;
      if (n == 1)
        cout << "a\n";
      else {
        if (n % 2 == 0) {
          for (long long int i = 0; i < n / 2; i++) cout << "a";
          cout << "b";
          for (long long int i = 0; i < (n / 2) - 1; i++) cout << "a";
          cout << "\n";
        } else {
          n--;
          for (long long int i = 0; i < n / 2; i++) cout << "a";
          cout << "b";
          for (long long int i = 0; i < (n / 2) - 1; i++) cout << "a";
          cout << "c\n";
        }
      }
      t--;
    }
    return 0;
  }
  