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 = 1000010;
const int M = 200010;
const long long inf = 0x7fffffff;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
int n, flag, ans;
int a[N];
void add(int x, int k) {
  while (x <= n) {
    a[x] += k;
    x += x & -x;
  }
}
int ask(int x) {
  int ret = 0;
  while (x) {
    ret += a[x];
    x -= x & -x;
  }
  return ret;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int tmp = read();
    add(tmp, 1);
    ans += ask(n) - ask(tmp);
  }
  if (n & 1) flag = 1;
  if (flag) {
    if (ans & 1)
      printf("Petr");
    else
      printf("Um_nik");
  } else {
    if (ans & 1)
      printf("Um_nik");
    else
      printf("Petr");
  }
  return 0;
}
