CodeForces 82A Double Cola

A. Double Cola

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.

For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.

Write a program that will print the name of a man who will drink the n-th can.

Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.

Input

The input data consist of a single integer n (1 ≤ n ≤ 109).

It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.

Output

Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.

Examples

input

Copy

1

output

Copy

Sheldon

input

Copy

6

output

Copy

Sheldon

input

Copy

1802

output

Copy

Penny

 题目大意：五个人排队买喝的，一个人买完后，会变成两个他在队尾继续排队，例如：a b c d e aa bb cc dd ee aaaa bbbb cccc dddd eeee……问你第你n个人是谁
思路:观察可得是个等比数列，第一个5，第二个10，第三个20，……公比是2，我们可以求出前n项和，找到这一轮是第几轮，再除以本轮的倍数即可，具体看代码（PS:貌似这题数据有点水，之前看到一篇博客上面n=7竟然输出的是Leonard，这样也能过！！！太水了。

/*
题目大意：五个人排队买喝的，一个人买完后，会变成两个他在队尾继续排队，问你第你n个人是谁
思路:观察可得是个等比数列，我们可以求出前n项和，找到这一轮是第几轮，再除以本轮的倍数即可，具体看代码
*/
#include<map>
#include<set>
#include <vector>
#include<stack>
#include<queue>
#include<cmath>
#include<string>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll unsigned long long
#define inf 0x3f3f3f
#define esp 1e-8
#define bug {printf("mmp\n");}
#define mm(a,b) memset(a,b,sizeof(a))
#define T() int test,q=1;scanf("%d",&test); while(test--)
const int maxn=2e5+10;
const double pi=acos(-1.0);
const int N=201;
const int mod=1e9+7;
ll a[maxn],b[maxn];
void ff()
{
    ll sum=5;
    for(int i=0; i<=100; i++)
    {
        a[i]=a[i-1]+sum;
        sum*=2;
       i>1? b[i]=b[i-1]*2:b[i]=2;
    }
}
void print(int n)
{
    switch(n)
    {
    case 1:
        cout<<"Sheldon"<<endl;
        break;
    case 2:
        cout<<"Leonard"<<endl;
        break;
    case 3:
        cout<<"Penny"<<endl;
        break;
    case 4:
        cout<<"Rajesh"<<endl;
        break;
    case 5:
        cout<<"Howard"<<endl;
        break;
    }
}
int main()
{
    ff();
    int n,index=0;
    cin>>n;
    if(n<6)
    {
        print(n);
        return 0;
    }
    for(int i=0; i<100; i++)
    {
        if(n<=a[i])
        {
            index=i;///找到他属于第几轮
            break;
        }
    }
    n-=a[index-1];
    int ans=(n-1)/b[index]+1;
    print(ans);
    return 0;
}