思维-B. Petr and Permutations

B. 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 11 to nn and then 3n3n 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+17n+1 times instead of 3n3n 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 nn (103≤n≤106103≤n≤106).

In the second line there are nn distinct integers between 11 and nn — the permutation of size nn from the test.

It is guaranteed that all tests except for sample are generated this way: First we choose nn — 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 nn) 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.

题意：给一个长度为 n 的序列 S表示，问 从 1-n 递增排序的序列 X表示变到 S（一次可以交换任意两个数字），总共交换的次数              如果是 3*n，则输出 Petr ，否则输出 另一个。

题解：找出逆序对，3*n - 逆序对数==偶数则输出Petr（易证）

#include
#include
#include
using namespace std;
int a[1000005],t[1000005],ans=0;

void dfs(int l,int r){
	if(l==r)
	  return ;
	int mid=(l+r)>>1;
	dfs(l,mid);
	dfs(mid+1,r);
	
	int i=l,j=mid+1,p=l;
	while(i<=mid&&j<=r){
		if(a[i]>a[j]){
			t[p++]=a[j++];
			ans+=(mid-i+1);
		}
		else{
			t[p++]=a[i++];
		}
	}
	while(i<=mid){
		t[p++]=a[i++];
	}
	while(j<=r){
		t[p++]=a[j++];
	}
	for(int k=l;k<=r;k++)
	   a[k]=t[k];
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=0;i