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 N = 1e6 + 4;
long long ans;
int a[N];
int tmp[N];
void merge_sort(int l, int r) {
  if (l >= r) return;
  int m = (l + r) / 2;
  merge_sort(l, m);
  merge_sort(m + 1, r);
  int cnt = 0;
  int j = m + 1;
  int i = l;
  int p = l, q = m + 1;
  int d = l;
  while (i <= m && j <= r) {
    if (a[i] > a[j])
      tmp[d++] = a[j++], ans += (m - i + 1);
    else
      tmp[d++] = a[i++];
  }
  while (i <= m) tmp[d++] = a[i++];
  while (j <= r) tmp[d++] = a[j++];
  for (int i = l; i <= r; ++i) a[i] = tmp[i];
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  merge_sort(1, n);
  if (ans % 2 == (3 * n) % 2)
    puts("Petr");
  else
    puts("Um_nik");
  return 0;
}
