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;
long long _mergeSort(long long arr[], long long temp[], long long left,
                     long long right);
long long merge(long long arr[], long long temp[], long long left,
                long long mid, long long right);
long long mergeSort(long long arr[], long long array_size) {
  long long *temp = (long long *)malloc(sizeof(long long) * array_size);
  return _mergeSort(arr, temp, 0, array_size - 1);
}
long long _mergeSort(long long arr[], long long temp[], long long left,
                     long long right) {
  long long mid, inv_count = 0;
  if (right > left) {
    mid = (right + left) / 2;
    inv_count = _mergeSort(arr, temp, left, mid);
    inv_count += _mergeSort(arr, temp, mid + 1, right);
    inv_count += merge(arr, temp, left, mid + 1, right);
  }
  return inv_count;
}
long long merge(long long arr[], long long temp[], long long left,
                long long mid, long long right) {
  long long i, j, k;
  long long inv_count = 0;
  i = left;
  j = mid;
  k = left;
  while ((i <= mid - 1) && (j <= right)) {
    if (arr[i] <= arr[j]) {
      temp[k++] = arr[i++];
    } else {
      temp[k++] = arr[j++];
      inv_count = inv_count + (mid - i);
    }
  }
  while (i <= mid - 1) temp[k++] = arr[i++];
  while (j <= right) temp[k++] = arr[j++];
  for (i = left; i <= right; i++) arr[i] = temp[i];
  return inv_count;
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void find(long long arrl[], long long exit[], long long n) {
  sort(arrl, arrl + n);
  sort(exit, exit + n);
  long long in = 1, max = 1, t = arrl[0];
  long long i = 1, j = 0;
  while (i < n && j < n) {
    if (arrl[i] <= exit[j]) {
      in++;
      if (in > max) {
        max = in;
        t = arrl[i];
      }
      i++;
    } else {
      in--;
      j++;
    }
  }
  cout << t << ' ';
}
int main() {
  unordered_map<long long, long long> mp;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long i, j, k, n, m, count = 0, sum = 0, x, y, z, c, a[1000000] = {0}, t,
                           ans;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  x = mergeSort(a, n);
  if (x % 2 == 0 && n % 2)
    cout << "Um_nik";
  else if (x % 2 == 0 && n % 2 == 0) {
    cout << "Petr";
  } else if (x % 2 && n % 2)
    cout << "Petr";
  else
    cout << "Um_nik";
}
