#include <bits/stdc++.h>
using namespace std;
using namespace chrono;
long long int ceil(long long int r, long long int d) {
  return (r / d) + (bool)(r % d);
}
void inputAArray(int siz, vector<int> &arr) {
  for (int i = 0; i < siz; i++) arr[i] = 1 + rand();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.length();
    if (n & 1) {
      cout << "NO" << endl;
      continue;
    }
    if (s.substr(0, n / 2) == s.substr(n / 2))
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
using namespace chrono;
long long int ceil(long long int r, long long int d) {
  return (r / d) + (bool)(r % d);
}
void inputAArray(int siz, vector<int> &arr) {
  for (int i = 0; i < siz; i++) arr[i] = 1 + rand();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.length();
    if (n & 1) {
      cout << "NO" << endl;
      continue;
    }
    if (s.substr(0, n / 2) == s.substr(n / 2))
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}

```