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;
  map<long long, long long> fib;
  bool isprime(int n) {
    int cnt = 0;
    for (int i = 1; i * i <= n; i++) {
      if (n % i == 0) {
        cnt += 2;
        if (i * i == n) {
          cnt--;
        }
      }
    }
    if (cnt == 2)
      return true;
    else
      return false;
  }
  int XOR(int n) {
    int ans;
    if (n % 4 == 0) {
      ans = 0;
    } else if (n % 2 == 0) {
      ans = 1;
    } else {
      if (abs(n - 1) % 4 == 0) {
        ans = n - 1;
      } else {
        ans = n;
      }
    }
    return ans;
  }
  bool isPowerOfTwo(long long n) {
    if (n == 0) return false;
    return (ceil(log2(n)) == floor(log2(n)));
  }
  void solve() {
    string second;
    cin >> second;
    if ((int)second.length() % 2 == 0) {
      int i = ((int)second.length() / 2) - 1;
      int j = (int)second.length() - 1;
      while (i >= 0) {
        if (second[i] == second[j]) {
          i--;
          j--;
        } else {
          cout << "NO\n";
          return;
        }
      }
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  