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.
#include <bits/stdc++.h>
using namespace std;
const int mx = 2e6 + 1000;
long long s[mx];
int main() {
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  long long sum = 0;
  for (int j = 0; j < 25; j++) {
    for (int i = 1; i <= n; i++) {
      if (s[i] != i) {
        long long k = s[s[i]];
        s[s[i]] = s[i];
        s[i] = k;
        sum++;
      }
    }
  }
  if (((7 * n + 1) % 2) == (sum % 2)) {
    cout << "Um_nik";
  } else {
    cout << "Petr";
  }
  return 0;
}
