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.*;
import java.io.*;
//import java.text.*;

public class Main{
    final long MOD = (long)1e9+7, IINF = (long)1e19;
    final int MAX = (int)500001, MX = (int)1e7+1, INF = (int)1e9;
//    DecimalFormat df = new DecimalFormat("0.00000000");
//    final double EPS = 1e-8;
    FastReader in;
    PrintWriter out;
    boolean multipleTC = false;
    
    public static void main(String[] args) throws Exception{new Main().run();};
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        for(int i = 1, t = (multipleTC)?ni():1; i<=t; i++)solve(i);
        out.flush();    
        out.close();
    }
    
    void solve(int TC){
        int n = ni();
        long[] a = la(0,n);
        long x = mergeSort(a, n);
        if(n%2==x%2)pn("Petr");
        else pn("Um_nik");
    }
    //Credits: https://www.geeksforgeeks.org/counting-inversions/
    static long mergeSort(long arr[], int array_size)
    {
        long temp[] = new long[array_size];
        return _mergeSort(arr, temp, 0, array_size - 1);
    }
     
    /* An auxiliary recursive method that sorts the input array and
      returns the number of inversions in the array. */
    static long _mergeSort(long arr[], long temp[], int left, int right)
    {
      int mid;
      long inv_count = 0;
      if (right > left)
      {
        /* Divide the array into two parts and call _mergeSortAndCountInv()
           for each of the parts */
        mid = (right + left)/2;
     
        /* Inversion count will be sum of inversions in left-part, right-part
          and number of inversions in merging */
        inv_count  = _mergeSort(arr, temp, left, mid);
        inv_count += _mergeSort(arr, temp, mid+1, right);
     
        /*Merge the two parts*/
        inv_count += merge(arr, temp, left, mid+1, right);
      }
      return inv_count;
    }
     
    /* This method merges two sorted arrays and returns inversion count in
       the arrays.*/
    static long merge(long arr[], long temp[], int left, int mid, int right)
    {
      int i, j, k;
      long inv_count = 0;
     
      i = left; /* i is index for left subarray*/
      j = mid;  /* j is index for right subarray*/
      k = left; /* k is index for resultant merged subarray*/
      while ((i <= mid - 1) && (j <= right))
      {
        if (arr[i] <= arr[j])
        {
          temp[k++] = arr[i++];
        }
        else
        {
          temp[k++] = arr[j++];
     
         /*this is tricky -- see above explanation/diagram for merge()*/
          inv_count = inv_count + (mid - i);
        }
      }
     
      /* Copy the remaining elements of left subarray
       (if there are any) to temp*/
      while (i <= mid - 1)
        temp[k++] = arr[i++];
     
      /* Copy the remaining elements of right subarray
       (if there are any) to temp*/
      while (j <= right)
        temp[k++] = arr[j++];
     
      /*Copy back the merged elements to original array*/
      for (i=left; i <= right; i++)
        arr[i] = temp[i];
     
      return inv_count;
    }
    
    long modPow(long a, long p){
        long o = 1;
        while(p>0){
            if(p%2==1)o = mul(o,a);
            a = mul(a,a);
            p>>=1;
        }
        return o;
    }
    
    long mul(long a, long b){
        if(a>=MOD)a%=MOD;
        if(b>=MOD)b%=MOD;
        return (a*b)%MOD;
    }
    
    int[] reverse(int[] a){
        int[] o = new int[a.length];
        for(int i = 0; i< a.length; i++)o[i] = a[a.length-i-1];
        return o;   
    }
    
    int[] sort(int[] a){
        if(a.length==1)return a;
        int mid = a.length/2;
        int[] b = sort(Arrays.copyOfRange(a,0,mid)), c = sort(Arrays.copyOfRange(a,mid,a.length));
        for(int i = 0, j = 0, k = 0; i< a.length; i++){
            if(j<b.length && k<c.length){
                if(b[j]<c[k])a[i] = b[j++];
                else a[i] = c[k++];
            }else if(j<b.length)a[i] = b[j++];
            else a[i] = c[k++];
        }
        return a;
    }
    
    long[] sort(long[] a){
        if(a.length==1)return a;
        int mid = a.length/2;
        long[] b = sort(Arrays.copyOfRange(a,0,mid)), c = sort(Arrays.copyOfRange(a,mid,a.length));
        for(int i = 0, j = 0, k = 0; i< a.length; i++){
            if(j<b.length && k<c.length){
                if(b[j]<c[k])a[i] = b[j++];
                else a[i] = c[k++];
            }else if(j<b.length)a[i] = b[j++];
            else a[i] = c[k++];
        }
        return a;
    }
    
    int[] ia(int ind,int n){
        int[] out = new int[ind+n];
        for(int i = 0; i< n; i++)out[ind+i] = ni();
        return out;
    }

    long[] la(int ind, int n){
        long[] out = new long[ind+n];
        for(int i = 0; i< n; i++)out[ind+i] = nl();
        return out;
    }

    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bitcount(long n){return (n==0)?0:(1+bitcount(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n(){return in.next();}
    String nln(){return in.nextLine();}
    int ni(){return Integer.parseInt(in.next());}
    long nl(){return Long.parseLong(in.next());}
    double nd(){return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }
            return str;
        }
    }
}