Codeforces Round #485 (Div. 2) E. Petr and Permutations

E. Petr and Permutations

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 $n$ (${10}^3\le n\le {10}^6$).

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

Copy

5
2 4 5 1 3

output

Copy

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.

分析：比赛的时候因为卡测评机了，导致D题提交了以后一直在in queue，再加上大晚上的困了，所以就没有开E做，结果第二天一看，这道题比D还容易啊，敲了一发提交就直接过了，看到这道题看到3n和7n+1之后随便带了几个数进去以后发现奇偶是不同的，每个序列交换的奇偶性肯定是  固定，然后就将大乱的序列重新交换位置得到原序列，用从cnt记录一下总共做了几次交换，然后直接比较cnt和3n的奇偶性，相同则输出Peter否则输出Um_nik。

AC代码：

#include
using namespace std;
const int maxn = 1000000+100;
int n;
int a[maxn];

int main(){
	cin>>n;
	for(int i=1; i<=n; i++){
		cin>>a[i];
	}
	int cnt = 0;
	for(int i=1; i<=n; i++){
		if(a[i] == i) continue;
		while(a[i] != i){
			cnt++;
			int t = a[i];
			swap(a[i], a[t]);
		}
	}
//	for(int i=1; i<=n; i++){
//		cout<