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 INF = 1e8;
int binpow(int a, int b) {
  if (b == 0) return a;
  if (b % 2)
    return a * binpow(a, b - 1);
  else {
    int x = binpow(a, b / 2);
    return x * x;
  }
}
long long n;
vector<long long> t;
int sum(int r) {
  long long result = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) result += t[r];
  return result;
}
void inc(long long i, long long delta) {
  for (; i < n; i = (i | (i + 1))) t[i] += delta;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  vector<long long> a(n);
  t.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    --a[i];
  }
  long long ans = 0;
  for (int i = n - 1; i >= 0; --i) {
    ans += sum(a[i]);
    inc(a[i], 1);
  }
  if (ans % 2 == (3 * n) % 2) {
    cout << "Petr";
  } else {
    cout << "Um_nik";
  }
  return 0;
}
