#include <bits/stdc++.h>
using namespace std;
const long long M = 1e5 + 10, OO = 2e9 + 10;
vector<char> v, vv;
long long n, m, k;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    v.clear();
    vv.clear();
    string s;
    cin >> s;
    bool f = 0;
    long long n = s.length();
    if (n & 1) {
      cout << "NO" << '\n';
      continue;
    }
    for (long long i = 0; i < n / 2; i++) v.push_back(s[i]);
    for (long long i = n / 2; i < n; i++) vv.push_back(s[i]);
    for (long long i = 0; i < v.size(); i++)
      if (v[i] != vv[i]) {
        f = 1;
        break;
      }
    if (f)
      cout << "NO" << '\n';
    else
      cout << "YES" << '\n';
  }
}

### 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;
const long long M = 1e5 + 10, OO = 2e9 + 10;
vector<char> v, vv;
long long n, m, k;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    v.clear();
    vv.clear();
    string s;
    cin >> s;
    bool f = 0;
    long long n = s.length();
    if (n & 1) {
      cout << "NO" << '\n';
      continue;
    }
    for (long long i = 0; i < n / 2; i++) v.push_back(s[i]);
    for (long long i = n / 2; i < n; i++) vv.push_back(s[i]);
    for (long long i = 0; i < v.size(); i++)
      if (v[i] != vv[i]) {
        f = 1;
        break;
      }
    if (f)
      cout << "NO" << '\n';
    else
      cout << "YES" << '\n';
  }
}

```