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 (103≤n≤106). 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.

給一個序列，是由1-n這個序列經過幾次任意兩個數交換得到的，如果交換次數是3n就是第一個人，如果是7n+1，就是第二個人 可以看出3n和7n+1一個是奇數一個是偶數，一開始想的是逆序數，猜逆序數個數奇偶性和n奇偶性若相同則第一個人，否則是第二個人 後來想到逆序數不行，考慮當前數列能交換幾次到原本的數列，然後判斷這個數的奇偶性再猜個結論

程式碼：

#include <bits/stdc++.h>
using namespace std;
const int N=1e6+50;
int a[N];
int n;
int main(void){
    scanf("%d",&n)
 
;
    int cnt=0;
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    for(int i=1;i<=n;i++){
        if(a[i]==i){
            continue;
        }
        else{
            while(a[i]!=i){
                cnt++;
                int t=a[i];
                swap(a[i],a[t]);
            }
 

        }
    }
    //printf("%d\n",ans);
    if(n%2==0 && cnt%2==0 || n%2==1 && cnt%2==1){
        printf("Petr\n");
    }
    else{
        printf("Um_nik\n");
    }
    return 0;
}