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 MAXN = 1001111;
int N;
int Num[MAXN];
int C[MAXN];
string Name[2] = {"Petr", "Um_nik"};
long long ANS = 0LL;
int lowbit(int k) { return k & (-k); }
void Add(int v) {
  for (int i = v; i <= N; i += lowbit(i)) ++C[i];
}
int Calc(int v) {
  int ret = 0;
  for (int i = v; i > 0; i -= lowbit(i)) ret += C[i];
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; ++i) cin >> Num[i];
  for (int i = N; i >= 1; --i) {
    ANS += Calc(Num[i] - 1);
    Add(Num[i]);
  }
  ANS &= 1;
  N &= 1;
  ANS ^= N;
  cout << Name[ANS] << endl;
  return 0;
}
