#include <bits/stdc++.h>
using namespace std;
long long sigma(long long l, long long r) {
  return ((r * (r + 1)) / 2) - ((l * (l - 1)) / 2);
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long dx[8] = {0, 0, 1, -1, 1, -1, 1, -1};
long long dy[8] = {1, -1, 0, 0, 1, -1, -1, 1};
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    string x;
    cin >> x;
    long long n = x.size() / 2;
    if (x.size() % 2)
      cout << "NO" << '\n';
    else {
      string s = x.substr(0, n);
      string d = x.substr(n, n);
      cout << (s == d ? "YES" : "NO") << '\n';
    }
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long sigma(long long l, long long r) {
  return ((r * (r + 1)) / 2) - ((l * (l - 1)) / 2);
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long dx[8] = {0, 0, 1, -1, 1, -1, 1, -1};
long long dy[8] = {1, -1, 0, 0, 1, -1, -1, 1};
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    string x;
    cin >> x;
    long long n = x.size() / 2;
    if (x.size() % 2)
      cout << "NO" << '\n';
    else {
      string s = x.substr(0, n);
      string d = x.substr(n, n);
      cout << (s == d ? "YES" : "NO") << '\n';
    }
  }
  return 0;
}

```