#include <bits/stdc++.h>
using namespace std;
int n;
char aa[1234][1234];
bool f[1234];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> aa[i];
  }
  for (int i = 0; i < n; i++) {
    if (strlen(aa[i]) % 2 == 1) {
      f[i] = true;
    } else {
      for (int j = 0; j < strlen(aa[i]) / 2; j++) {
        if (aa[i][j] != aa[i][strlen(aa[i]) / 2 + j]) {
          f[i] = true;
          break;
        }
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (f[i]) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int n;
char aa[1234][1234];
bool f[1234];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> aa[i];
  }
  for (int i = 0; i < n; i++) {
    if (strlen(aa[i]) % 2 == 1) {
      f[i] = true;
    } else {
      for (int j = 0; j < strlen(aa[i]) / 2; j++) {
        if (aa[i][j] != aa[i][strlen(aa[i]) / 2 + j]) {
          f[i] = true;
          break;
        }
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (f[i]) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
    }
  }
  return 0;
}

```