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.
import java.util.Scanner;

public class E {
static final int INF = Integer.MAX_VALUE;
	static int counter;
	static void mergeSort(int[] a, int b, int e)
	{
		if(b < e)
		{
			int q = (b + e) / 2;
			mergeSort(a, b, q);
			mergeSort(a, q + 1, e);
			merge(a, b, q, e);
		}
	}


    static void merge(int[] a, int b, int mid, int e)
    {
       int n1 = mid - b + 1;
       int n2 = e - mid;
       int[] L = new int[n1+1], R = new int[n2+1];
       
       for(int i = 0; i < n1; i++)  L[i] = a[b + i];
       for(int i = 0; i < n2; i++)  R[i] = a[mid + 1 + i];
       L[n1] = R[n2] = INF;
       
       for(int k = b, i = 0, j = 0; k <= e; k++)
    	   if(L[i] <= R[j])
    		   a[k] = L[i++];
    	   else {
    		   a[k] = R[j++];
    		   counter+=(L.length-i-1);
    	   }
    }  

public static void main(String[] args) {
	Scanner sc= new Scanner(System.in);
	int n = sc.nextInt();
	int [] elm =new int[n];
	for (int i = 0; i < elm.length; i++) {
		elm[i] = sc.nextInt();
	}
	mergeSort(elm, 0, n-1);
	int petr = 3*n;
	int alex = 7*n+1;
	petr-=counter;
	if(petr%2==0)
		System.out.println("Petr");
	else
		System.out.println("Um_nik");
}
}
