#include <bits/stdc++.h>
using namespace std;
long long int t, n, i, l, k, j;
void out() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int m = 0;
  char s[105];
  cin >> s;
  l = strlen(s);
  if (l % 2 == 1) {
    cout << "NO" << endl;
  } else {
    k = (l / 2) - 1;
    for (i = 0, j = k + 1; i <= k, j <= l - 1; i++, j++) {
      if (s[i] != s[j]) {
        m++;
        break;
      }
    }
    if (m == 0) {
      cout << "YES" << endl;
      ;
    } else {
      cout << "NO" << endl;
      ;
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> t;
  while (t--) {
    out();
  }
  return 0;
}

### Prompt

Your task is to create 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 int t, n, i, l, k, j;
void out() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int m = 0;
  char s[105];
  cin >> s;
  l = strlen(s);
  if (l % 2 == 1) {
    cout << "NO" << endl;
  } else {
    k = (l / 2) - 1;
    for (i = 0, j = k + 1; i <= k, j <= l - 1; i++, j++) {
      if (s[i] != s[j]) {
        m++;
        break;
      }
    }
    if (m == 0) {
      cout << "YES" << endl;
      ;
    } else {
      cout << "NO" << endl;
      ;
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> t;
  while (t--) {
    out();
  }
  return 0;
}

```