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
from sys import stdin
t = int(stdin.readline())
for _ in xrange(t):
	n = int(stdin.readline())
	a = map(int,stdin.readline().split())
	di = {}
	for i in xrange(n):
		di[a[i]] = i + 1
	valid = 1
	#print di
	for i in xrange(n):
		if a[i]  !=di[i+1]:
			valid = 0
	if valid:
		print "inverse"
	else:
		print "not inverse"