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 cnt = 0;
long int a[1000010];
int poi1;
int poi2;
int merge(int l, int r) {
  if (l == r - 1) return 0;
  int mid = (l + r) / 2;
  merge(l, mid);
  merge(mid, r);
  int b[r - l];
  poi1 = l;
  poi2 = mid;
  for (int i = 0; i < r - l; i++) {
    if (poi1 >= mid) {
      b[i] = a[poi2++];
    } else if (poi2 >= r) {
      b[i] = a[poi1++];
      cnt += poi2 - mid;
    } else if (a[poi1] == a[poi2]) {
      b[i] = a[poi1++];
      cnt += poi2 - mid;
    } else if (a[poi1] < a[poi2]) {
      b[i] = a[poi1++];
      cnt += poi2 - mid;
    } else {
      b[i] = a[poi2++];
    }
  }
  for (int i = 0; i < r - l; i++) {
    a[i + l] = b[i];
  }
  return 0;
}
int main() {
  long long int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  merge(0, n);
  if (n % 2 == 1) {
    if (cnt % 2 == 1) {
      cout << "Petr";
    } else
      cout << "Um_nik";
  } else {
    if (cnt % 2 == 0) {
      cout << "Petr";
    } else
      cout << "Um_nik";
  }
}
