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;
int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }
long long n, tree[4040404], s = 1, a;
pair<long long, long long> d[1010101];
void update(int x, int y) {
  x += s - 1;
  tree[x] = y;
  while (x /= 2) tree[x] = tree[x * 2] + tree[x * 2 + 1];
}
long long sum(int x, int y) {
  x += s - 1;
  y += s - 1;
  long long ans = 0;
  while (x <= y) {
    if (x % 2 == 1) ans += tree[x++];
    if (y % 2 == 0) ans += tree[y--];
    if (x > y) break;
    x /= 2;
    y /= 2;
  }
  return ans;
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &d[i].first);
    d[i].second = i;
  }
  while (s < n) s *= 2;
  sort(d + 1, d + n + 1);
  for (long long i = 1; i <= n; i++) {
    update(d[i].second, 1);
    a += sum(d[i].second + 1, n);
  }
  if (n % 2 == 1) {
    if (a % 2 == 1)
      puts("Petr");
    else
      puts("Um_nik");
  } else {
    if (a % 2 == 1)
      puts("Um_nik");
    else
      puts("Petr");
  }
}
