Let num (> 0) be n (> 0) digit(s) positive integer. num is represented as N₁N₂N₃N₄.....N_n-2N_n-1N_n , where N_i is the i^th digit of num from left (0 < i < n+1). Digits of num are sorted in descending and ascending order respectively and this sorting generates two new positive integers num_dsc and num_asc. The difference between the numbers is diff_num = num_dsc - num_asc , if diff_num is divisible by both 6 and 9, then we say that num is a magic number. Let sum_digits is defined as following

number = diff_num
do {
number = sum of digits of number
 } while (number > 10)

 sum_digits = number

Input

First line of input is t (< 101), total number of test cases. Each test case has n (< 10001) as its first input and next n lines contains num (< 10¹⁰⁰). 

Output

For each test case, write exactly n lines containing two/three specifications without space :
(i) Y if num is magic number otherwise N.
(ii) Let sum_digits = c, ZER if c is 0 (zero), ONE if c is 1 (one) if c > 1, EP if c is even and prime, ENP if c is even but not prime, OP if c is odd and prime or ONP if c is odd but not prime.
(iii) Let diff_num = d, If num is not a magic number then print EQL if d is not divisible by both 6 and 9, LTN if d is not divisible by 6 only, GTN if d is not divisible by 9 only.

Example

Input:
1
2
31
100

Output:
YONP
NONPLTN

0 is divisible by 6 and 9 :)