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
T = int(raw_input())

while(T):
	T = T - 1
	
	n = int(raw_input())
	
	num = [int(c) for c in raw_input().split()]
	num2 = []
	for each in num:
		num2.append(each)
	flag = 0
	
	for i in range(0,len(num2)):
		t = num2[i]
		if(num2[i] > n):
			flag = 1
			break
		num2[i] = num2[num2[i]-1]
		num2[num2[i]-1] = t
	
	for i in range(0,len(num)):
		if(num[i] != num2[i]):
			flag = 1
			break
	if(flag == 1):
		print "not inverse"
	else:
		print "inverse"