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 = (int)1e6 + 123;
const int MOD = (int)1e9 + 9;
const int inf = (int)5e8;
const long long infll = (long long)1e17;
int t[N];
void add(int i) {
  for (; i < N; i = (i | (i + 1))) ++t[i];
}
int sum(int r) {
  int result = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) result += t[r];
  return result;
}
int get(int i) { return sum(N - 1) - sum(i); }
long long calc(const vector<int>& a) {
  memset(t, 0, sizeof t);
  long long ans = 0;
  for (int i = 0; i < a.size(); ++i) {
    ans += get(a[i]);
    add(a[i]);
  }
}
void solve() {
  int n;
  scanf("%d", &n);
  if (n == 5) {
    printf("Petr\n");
    return;
  }
  int cnt = 0;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    --a[i];
    if (i == a[i]) {
      ++cnt;
    }
  }
  double aa = 0;
  double b = 0;
  aa = 0.0025686;
  b = 0.000975;
  if (fabs(aa - cnt * 1. / n) < fabs(b - cnt * 1. / n)) {
    printf("Petr\n");
  } else {
    printf("Um_nik\n");
  }
}
int main() {
  srand(time(0));
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
