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;
long long int minswaps(long long int arr[], long long int n) {
  pair<int, int> arrpos[n];
  for (long long int i = 0; i < n; i++) {
    arrpos[i].first = arr[i];
    arrpos[i].second = i;
  }
  sort(arrpos, arrpos + n);
  vector<bool> vis(n, false);
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    if (vis[i] || arrpos[i].second == i) continue;
    long long int cycle_size = 0;
    int j = i;
    while (!vis[j]) {
      vis[j] = true;
      j = arrpos[j].second;
      cycle_size++;
    }
    ans += (cycle_size - 1);
  }
  return ans;
}
int main() {
  long long int n;
  cin >> n;
  long long int arr[n];
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  long long int swaps = minswaps(arr, n);
  if ((((3 * n) - swaps) % 2) == 0)
    cout << "Petr";
  else
    cout << "Um_nik";
}
