# Petr and Permutations

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

52 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.

Solution:

 123456789101112131415161718192021222324252627282930 #include using namespace std; int main() {   ios::sync_with_stdio(false);   cin.tie(0);   int n;   cin >> n;   vector a(n);   for (int i = 0; i < n; i++) {     cin >> a[i];     a[i]--;   }   vector was(n, 0);   int cyc = 0;   for (int i = 0; i < n; i++) {     if (was[i]) {       continue;     }     cyc++;     int p = i;     while (!was[p]) {       was[p] = 1;       p = a[p];     }   }   cout << ((n - cyc) % 2 == (3 * n) % 2 ? "Petr" : "Um_nik") << '\n';   return 0; }