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;
vector<long long> fenv(1000000);
long long n;
void addfenv(int p, int x) {
  for (; p < n; p |= p + 1) fenv[p] += x;
}
long long getfenv(int r) {
  long long res = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) res += fenv[r];
  return res;
}
int main() {
  cin >> n;
  vector<long long> arr(n);
  for (auto &i : arr) cin >> i, --i;
  long long cnt = 0;
  for (int i = n - 1; i >= 0; --i) cnt += getfenv(arr[i]), addfenv(arr[i], 1);
  if (cnt % 2 == n % 2)
    cout << "Petr";
  else
    cout << "Um_nik";
  return 0;
}
