Petr likes to come up with problems about randomly generated data. This time problem is about random permutation. He decided to generate a random permutation this way: he takes identity permutation of numbers from 1 to n and then 3n times takes a random pair of different elements and swaps them. Alex envies Petr and tries to imitate him in all kind of things. Alex has also come up with a problem about random permutation. He generates a random permutation just like Petr but swaps elements 7n+1 times instead of 3n times. Because it is more random, OK?!

You somehow get a test from one of these problems and now you want to know from which one.

Input

In the first line of input there is one integer n (10^{3} ≤ n ≤ 10^{6}).

In the second line there are n distinct integers between 1 and n — the permutation of size n from the test.

It is guaranteed that all tests except for sample are generated this way: First we choose n — the size of the permutation. Then we randomly choose a method to generate a permutation — the one of Petr or the one of Alex. Then we generate a permutation using chosen method.

Output

If the test is generated via Petr's method print "Petr" (without quotes). If the test is generated via Alex's method print "Um_nik" (without quotes).

Example

Input

5
2 4 5 1 3


Output

Petr

Note

Please note that the sample is not a valid test (because of limitations for n) and is given only to illustrate input/output format. Your program still has to print correct answer to this test to get AC.

Due to randomness of input hacks in this problem are forbidden.
n=input()
l=map(int, raw_input().split())
odd = False
if n%2 == 1:
	n += 1
	odd = True
	l.append(n)
d={}
ev=[]
od=[]
c=0
for i in l:
	d[i]=c
	c+=1
	if i&1:
		od.append(i)
	else:
		ev.append(i)
ev.sort()
od.sort()
pr=[]
for i in xrange(n/2):
	pr.append(od[i])
	pr.append(ev[i])
#print pr
c=0
for i in xrange(n):
	if pr[i]!=l[i]:
		z=d[pr[i]]
		tm=l[i]
		l[i]=l[z]
		l[z]=tm
		d[l[z]]=z
		c+=1
if not odd:
	if c%2 == 0:
		print "Petr"
	else:
		print "Um_nik"
else:
	if c%2 == 0:
		print "Um_nik"
	else:
		print "Petr"
	