#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e6;
void solve() {
  string s;
  cin >> s;
  long long n = s.size();
  if (n % 2) {
    cout << "NO" << '\n';
    return;
  }
  bool ans = 1;
  for (long long i = 0; i < n / 2; i++) {
    if (s[i] != s[i + (n / 2)]) {
      ans = 0;
      break;
    }
  }
  if (ans)
    cout << "YES" << '\n';
  else
    cout << "NO" << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long num = 1;
  long long _t;
  cin >> _t;
  for (long long i = 1; i <= _t; i++) {
    solve();
  }
}

### Prompt

Your challenge is to write 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;
const long long maxn = 1e6;
void solve() {
  string s;
  cin >> s;
  long long n = s.size();
  if (n % 2) {
    cout << "NO" << '\n';
    return;
  }
  bool ans = 1;
  for (long long i = 0; i < n / 2; i++) {
    if (s[i] != s[i + (n / 2)]) {
      ans = 0;
      break;
    }
  }
  if (ans)
    cout << "YES" << '\n';
  else
    cout << "NO" << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long num = 1;
  long long _t;
  cin >> _t;
  for (long long i = 1; i <= _t; i++) {
    solve();
  }
}

```