#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    string p, q;
    long long int count = 0;
    for (long long int i = 0; s[i] != '\0'; i++) count++;
    if (count == 1)
      cout << "no" << endl;
    else {
      p = s[0];
      for (long long int i = 1; i < (count / 2); i++) {
        p += s[i];
      }
      q = s[count / 2];
      for (long long int i = (count / 2) + 1; i < count; i++) {
        q += s[i];
      }
      if (p == q)
        cout << "yes" << endl;
      else
        cout << "no" << endl;
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    string p, q;
    long long int count = 0;
    for (long long int i = 0; s[i] != '\0'; i++) count++;
    if (count == 1)
      cout << "no" << endl;
    else {
      p = s[0];
      for (long long int i = 1; i < (count / 2); i++) {
        p += s[i];
      }
      q = s[count / 2];
      for (long long int i = (count / 2) + 1; i < count; i++) {
        q += s[i];
      }
      if (p == q)
        cout << "yes" << endl;
      else
        cout << "no" << endl;
    }
  }
  return 0;
}

```