B'Oh, that 's funny, is it? Oh it 's funny? Because we 've been at this for twelve hours and you haven 't solved it either, so I don 't know why you 're laughing. You 've got one hour! Solve it! Wheatley decided to try to make a test chamber. He made a nice test chamber, but there was only one detail absent -- cubes. For completing the chamber Wheatley needs n cubes. i -th cube has a volume a_i . Wheatley has to place cubes in such a way that they would be sorted in a non-decreasing order by their volume. Formally, for each i>1 , a_{i-1} <= a_i must hold. To achieve his goal, Wheatley can exchange two neighbouring cubes. It means that for any i>1 you can exchange cubes on positions i-1 and i . But there is a problem: Wheatley is very impatient. If Wheatley needs more than frac{n cdot (n-1)}{2}-1 exchange operations, he won 't do this boring work. Wheatly wants to know: can cubes be sorted under this conditions? Each test contains multiple test cases. The first line contains one positive integer t ( 1 <= t <= 1000 ), denoting the number of test cases. Description of the test cases follows. The first line of each test case contains one positive integer n ( 2 <= n <= 5 cdot 10^4 ) -- number of cubes. The second line contains n positive integers a_i ( 1 <= a_i <= 10^9 ) -- volumes of cubes. It is guaranteed that the sum of n over all test cases does not exceed 10^5 . For each test case, print a word in a single line: "YES" (without quotation marks) if the cubes can be sorted and "NO" (without quotation marks) otherwise. In the first test case it is possible to sort all the cubes in 7 exchanges. In the second test case the cubes are already sorted. In the third test case we can make 0 exchanges, but the cubes are not sorted yet, so the answer is "NO". '...