Solve in Python:
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 $n$ ($10^{3} \le n \le 10^{6}$).

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 = int(input())
l = list(map(int, input().split()))
c = 1
ans = 0
vis = [0] * 1000005
for i in range(n):
    p = i
    vis[i + 1] = 1
    while vis[l[p]] != 1:
        ans -= -1
        vis[l[p]] = 1
        p = l[p] - 1
if 3 * n % 2 == ans % 2: print("Petr")
else: print("Um_nik")