description: A string is called square if it is some string written twice in a row. For example, the strings "aa", "abcabc", "abab" and "baabaa" are square. But the strings "aaa", "abaaab" and "abcdabc" are not square.
  
  For a given string s determine if it is square.
  
  Input
  
  The first line of input data contains an integer t (1 ≤ t ≤ 100) —the number of test cases.
  
  This is followed by t lines, each containing a description of one test case. The given strings consist only of lowercase Latin letters and have lengths between 1 and 100 inclusive.
  
  Output
  
  For each test case, output on a separate line:
  
    * YES if the string in the corresponding test case is square, 
    * NO otherwise. 
  
  
  
  You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive response).
  
  Example
  
  Input
  
  
  10
  a
  aa
  aaa
  aaaa
  abab
  abcabc
  abacaba
  xxyy
  xyyx
  xyxy
  
  
  Output
  
  
  NO
  YES
  NO
  YES
  YES
  YES
  NO
  NO
  NO
  YES
solution: #include <bits/stdc++.h>
  using namespace std;
  template <typename A, typename B>
  ostream &operator<<(ostream &os, const pair<A, B> &p) {
    return os << '(' << p.first << ", " << p.second << ')';
  }
  template <typename T_container, typename T = typename enable_if<
                                      !is_same<T_container, string>::value,
                                      typename T_container::value_type>::type>
  ostream &operator<<(ostream &os, const T_container &v) {
    os << '{';
    string sep;
    for (const T &x : v) os << sep << x, sep = ", ";
    return os << '}';
  }
  void dbg_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cerr << ' ' << H;
    dbg_out(T...);
  }
  const int MAX_N = 1e5 + 5;
  const long long MOD = 1e9 + 7;
  const long long INF = 1e9;
  const long double EPS = 1e-9;
  void solve() {
    string str;
    cin >> str;
    int len = ((int)str.size());
    if (len & 1)
      cout << "NO" << endl;
    else {
      for (int i = 0; i < len / 2; ++i) {
        if (str[i] != str[i + len / 2]) {
          cout << "NO" << endl;
          return;
        }
      }
      cout << "YES" << endl;
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int tc = 1;
    cin >> tc;
    for (int t = 1; t <= tc; t++) {
      solve();
    }
  }
  