#include <bits/stdc++.h>
using namespace std;
void solve() {
  string s;
  cin >> s;
  int n = s.size();
  string answer = "";
  if (n & 1 ^ 1) {
    string pref, suff;
    pref = suff = "";
    for (int i = 0; i < n / 2; i++) pref.push_back(s[i]);
    for (int i = n / 2; i < n; i++) suff.push_back(s[i]);
    if (pref == suff)
      answer = "YES";
    else
      answer = "NO";
  } else
    answer = "NO";
  cout << answer << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
void solve() {
  string s;
  cin >> s;
  int n = s.size();
  string answer = "";
  if (n & 1 ^ 1) {
    string pref, suff;
    pref = suff = "";
    for (int i = 0; i < n / 2; i++) pref.push_back(s[i]);
    for (int i = n / 2; i < n; i++) suff.push_back(s[i]);
    if (pref == suff)
      answer = "YES";
    else
      answer = "NO";
  } else
    answer = "NO";
  cout << answer << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```