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 power(long long a, long long b, long long m) {
    long long res = 1;
    a = a % m;
    if (a == 0) {
      return 0;
    }
    while (b > 0) {
      if (1 & b) {
        res = (res * a) % m;
      }
      b = b >> 1;
      a = (a * a) % m;
    }
    return res;
  }
  bool cmp(long long a, long long b) {
    if (a > b) {
      return true;
    }
    return false;
  }
  bool cmpi(pair<long long, long long> a, pair<long long, long long> b) {
    if (a.first > b.first) {
      return true;
    } else if (a.first == b.first) {
      if (a.second >= b.second) {
        return true;
      } else {
        return false;
      }
    }
    return false;
  }
  long long gcd(long long a, long long b) {
    if (a < b) {
      swap(a, b);
    }
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  void solve() {
    string s;
    cin >> s;
    long long n = s.size(), i;
    if (n % 2 == 1) {
      cout << "NO";
      return;
    }
    long long m = n / 2;
    for (i = 0; i < m; i++) {
      if (s[i] != s[i + m]) {
        cout << "NO";
        return;
      }
    }
    cout << "YES";
    return;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      solve();
      cout << endl;
    }
  }
  