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;
struct fenwick {
  long long n, fen[1000005];
  void resize(int _n) {
    n = _n + 1;
    memset(fen, 0, sizeof fen);
  }
  long long query(int ind) {
    long long ans = 0;
    while (ind >= 1) {
      ans += fen[ind];
      ind -= ind & -ind;
    }
    return ans;
  }
  void update(int ind, long long val) {
    while (ind <= n) {
      fen[ind] += val;
      ind += ind & -ind;
    }
  }
  long long query(int l, int r) { return query(r) - query(l - 1); }
};
int n, x;
long long inv;
vector<int> a;
fenwick tree;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 0; i < n && cin >> x; i++) a.push_back(x);
  tree.resize(n + 1);
  for (int i = 0; i < a.size(); i++) {
    inv += tree.query(a[i] + 1, n);
    tree.update(a[i], 1);
  }
  if (inv % 2 == 3 * n % 2)
    cout << "Petr" << endl;
  else
    cout << "Um_nik" << endl;
  return 0;
}
