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>
#pragma comment(linker, "/STACK:836777216")
using namespace std;
int a[1000005];
bool used[1000005];
int n;
int cnt = 0;
int main(void) {
  scanf("%d", &n);
  for (int(i) = 0; (i) < (n); (i)++) {
    scanf("%d", &a[i]);
  }
  memset((used), (0), sizeof((used)));
  for (int(i) = 0; (i) < (n); (i)++) {
    if (used[i]) {
      continue;
    }
    if (a[i] == i + 1) {
      cnt++;
      used[i] = 1;
      continue;
    }
    int cur = a[i] - 1;
    while (!used[cur]) {
      used[cur] = 1;
      cur = a[cur] - 1;
    }
    cnt++;
  }
  if (cnt % 2) {
    printf("Um_nik\n");
  } else {
    printf("Petr\n");
  }
}
