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 d[1000000 + 5], a[1000000 + 5];
int n;
int lowbit(int x) { return x & (-x); }
void upd(int x, int val) {
  while (x <= n) {
    d[x] += val;
    x += lowbit(x);
  }
}
int query(int x) {
  int sum = 0;
  while (x) {
    sum += d[x];
    x -= lowbit(x);
  }
  return sum;
}
int main() {
  while (scanf("%d", &n) != EOF && n) {
    int i, x;
    memset(d, 0, sizeof(d));
    memset(a, 0, sizeof(a));
    long long ans = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      upd(a[i], 1);
      ans += (i - query(a[i]));
    }
    int judge = 3 * n;
    if (ans % 2 == judge % 2)
      printf("Petr\n");
    else
      printf("Um_nik\n");
  }
}
