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 long long N = 1e6 + 6;
long long tree[4 * N], a[N];
pair<long long, long long> b[N];
void update(long long nd, long long beg, long long en, long long idx) {
  if (beg > idx || en < idx) return;
  if (beg == en && beg == idx) {
    tree[nd]++;
    return;
  }
  long long mid = (beg + en) / 2;
  update(nd * 2, beg, mid, idx);
  update(nd * 2 + 1, mid + 1, en, idx);
  tree[nd] = tree[nd * 2] + tree[nd * 2 + 1];
  return;
}
long long query(long long nd, long long beg, long long en, long long l,
                long long r) {
  if (beg > r || en < l) return 0;
  if (beg >= l && en <= r) return tree[nd];
  long long mid = (beg + en) / 2;
  long long p1 = query(nd * 2, beg, mid, l, r);
  long long p2 = query(nd * 2 + 1, mid + 1, en, l, r);
  return p1 + p2;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n = 0, ans = 0;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = {a[i], i};
  }
  sort(b + 1, b + n + 1, greater<pair<long long, long long> >());
  for (long long i = 1; i <= n; i++) {
    long long val = b[i].first, idx = b[i].second;
    ans += query(1, 1, n, 1, idx);
    update(1, 1, n, idx);
  }
  ans %= 2;
  n %= 2;
  if (ans == n)
    cout << "Petr\n";
  else
    cout << "Um_nik\n";
  return 0;
}
