## PSYCHOT - Psycho34 (easy)

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In the prime factorization of a number N, there's two kinds of powers. Even powers, in red, are psychotic ones, and odd powers, in blue, are ordinary ones.
We'll say N a Psycho number if the count of even powers is strictly greater than the count of odd powers, else an Ordinary number.
For example, if N = 67500 with prime factorization 67500 = 22 x 33 x 54 .
This number have 2 even powers and 1 odd power. Since 2>1, so the number 67500 is a Psycho Number.

### Input

The first line of input contains an integer T, the number of test cases.

Each of the next T lines contains one integer N.

### Output

For each test case, print if N is Psycho or Ordinary number.

### Example

```Input:
2
3
4

Output:
Ordinary Number
Psycho Number

```

### Constraints

```0 < T < 10^4
1 < N < 10^14
```

Time limit is ×2 my top speed with Python3 language, it could be not easy with slow languages.
O(N^.5 / log(N)) should give TLE even with fast languages. You are awaited to submit something between O(N^0.33 / log(N)) and O(N^0.25 / log(N)). You can try before the quite similar "tutorial" problem : Psycho before.
@speed addicts : my top C timing is 0.02s. (TL updated 2017-02-11 ; compiler changes)