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
import re
import string

t=int(input())
mylist=[]

def inv (l, digits):
	newdigit=[]
	for i in range(l):
		p=digits[i]
		newp=digits[p-1]
		newdigit.insert(newp-1,p)
	#print(newdigit)
	if cmp(newdigit, digits):
		return('not inverse')
	else :
		return('inverse')




for i in range(t):
	n=int(input())
	
	#for i in range(n):

	no=raw_input()

	mylist=(no.split())
#print(mylist)
	for i in range(len(mylist)):
		mylist[i]=string.atoi(mylist[i])

#print(mylist)
	print(inv(n,mylist))