There are many ways to order a list of integers from 1 to n. For example, if  n = 3, the list could be : [3\; 1\; 2].

But there is a special way to create another list from the given list of integers. In this list, position of integer i is the i-th number in the given list. So following this rule, the given list will be written as: [2\; 3\; 1]. This list is called inverse list. Now there exists some list whose inverse list is identical. For example, inverse list of [1\; 2\; 3] is same as given list. Given a list of integers you have to determine whether the list is inverse or not.

The input contains several test cases. The first line is the number of test cases t (1 ≤ t ≤ 100) . The first line of each test case contains an integer n \;(1 ≤ n ≤ 100000). Then a list of the integers 1 to n follows in the next line. 

SAMPLE INPUT
2
3
3 1 2
3
1 2 3

SAMPLE OUTPUT
not inverse
inverse
n=int(raw_input())
while n:
	n=n-1
	size=int(raw_input())
	a=raw_input().split()
	a=[int(i) for i in a]
	b=list(xrange(size))
	for i in list(xrange(size)):
		b[a[i]-1]=i+1
	
	if a==b:
		print "inverse"
	else:
		print "not inverse"