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;
  long long int power(long long int x, long long int y) {
    if (y == 0)
      return 1;
    else {
      long long int result = power(x, y / 2) % 1000000007;
      if (y % 2 == 0)
        return (result % 1000000007 * result % 1000000007) % 1000000007;
      else
        return (result % 1000000007 * result % 1000000007 * x % 1000000007) %
               1000000007;
    }
  }
  long long int gcd(long long int a, long long int b) {
    if (b == 0)
      return a;
    else
      return gcd(b, a % b);
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int t = 1;
    cin >> t;
    while (t--) {
      string s;
      cin >> s;
      if (s.length() % 2 != 0) {
        cout << "NO" << '\n';
      } else {
        int n = s.length() / 2;
        string s1, s2;
        int cnt = 0;
        for (int i = 0; i < n; i++) {
          if (s[i] == s[i + n])
            cnt++;
          else
            break;
        }
        if (cnt == n) {
          cout << "YES" << '\n';
        } else
          cout << "NO" << '\n';
      }
    }
    return 0;
  }
  