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 = 1000005;
long long mod = 1000000009;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
bool is_vowel(char c) {
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'y')
    return 1;
  return 0;
}
long double getDistance(long double x1, long double y1, long double x2,
                        long double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
long long extended_euclidean(long long a, long long b, long long &x,
                             long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long g = extended_euclidean(b, a % b, y, x);
  y -= (a / b) * x;
  return g;
}
long long power(long long base, long long p) {
  if (p == 1) return base;
  if (!p) return 1ll;
  long long ret = power(base, p / 2);
  ret *= ret;
  ret %= mod;
  if (p & 1) ret *= base;
  return ret % mod;
}
pair<int, int> vec(pair<int, int> a, pair<int, int> b) {
  pair<int, int> ret = pair<int, int>(a.first - b.first, a.second - b.second);
  return ret;
}
long long getArea(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
  pair<int, int> x = vec(a, b);
  pair<int, int> y = vec(c, b);
  return abs(x.first * y.second - x.second * y.first);
}
int n, arr[N];
vector<int> adj[N];
bool v[N];
int cnt = 0;
void dfs(int node) {
  if (v[node]) return;
  cnt++;
  v[node] = 1;
  for (int i = (0); i <= (int(adj[node].size()) - 1); ++i) dfs(adj[node][i]);
}
int main() {
  cin >> n;
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &arr[i]);
    adj[arr[i]].push_back(i);
    adj[i].push_back(arr[i]);
  }
  int sum = 0;
  for (int i = (1); i <= (n); ++i) {
    if (v[i]) continue;
    cnt = 0;
    dfs(i);
    sum += cnt - 1;
  }
  int way1 = 3 * n - sum;
  int way2 = 7 * n + 1 - sum;
  if (way2 % 2 == 0)
    cout << "Um_nik";
  else
    cout << "Petr";
  return 0;
}
