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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  map<int, int> a;
  int b[n], c[n];
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
    pair<int, int> s;
    s = make_pair(i + 1, i + 1);
    a.insert(s);
    c[i] = i + 1;
  }
  map<int, int>::iterator itr1, itr2;
  int count = 0;
  for (int i = 0; i < n; ++i) {
    if (b[i] != c[i]) {
      count++;
      itr1 = a.find(b[i]);
      itr2 = a.find(c[i]);
      int k = itr1->second;
      a.erase(itr1);
      pair<int, int> s = make_pair(c[i], k);
      c[k - 1] = c[i];
      a.erase(itr2);
      a.insert(s);
    }
  }
  if (n % 2 == 0) {
    if (count % 2 == 0)
      cout << "Petr\n";
    else
      cout << "Um_nik\n";
  } else {
    if (count % 2 == 1)
      cout << "Petr\n";
    else
      cout << "Um_nik\n";
  }
}
