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 FenwickTree {
  vector<int> v;
  FenwickTree(int n) : v(n, 0) {}
  void add(int i, int x) {
    for (; i < (int)v.size(); i |= i + 1) v[i] += x;
  }
  int sum(int i) {
    int r = 0;
    for (--i; i >= 0; i = (i & (i + 1)) - 1) r += v[i];
    return r;
  }
  int sum(int left, int right) { return sum(right) - sum(left); }
};
long long inversionNumber(int x[], int n) {
  int *c = new int[n];
  memcpy(c, x, sizeof(*x) * n);
  sort(c, c + n);
  int m = unique(c, c + n) - c;
  FenwickTree ft(m);
  long long res = 0;
  for (int i = n - 1; i >= 0; i--) {
    int idx = lower_bound(c, c + m, x[i]) - c;
    res += ft.sum(idx);
    ft.add(idx, 1);
  }
  delete[] c;
  return res;
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    vector<int> perm(n);
    for (int i = 0; i < n; ++i) scanf("%d", &perm[i]), --perm[i];
    int parity = inversionNumber(perm.data(), n) % 2;
    puts(parity == n % 2 ? "Petr" : "Um_nik");
  }
}
