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

1


Output

Sheldon


Input

6


Output

Sheldon


Input

1802


Output

Penny
#-*-coding:utf-8-*-

import string
import sys
reload(sys)
sys.setdefaultencoding('utf-8')


# 하나의 큐
# 초기 큐상태 Sheldon, Leonard, Penny, Rajesh and Howard 
# 하나를 먹으면 먹은 사람을 큐에 2개 추가

# Penny 가 콜라 3개를 먹음
# Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.


n = input()


name = ['Sheldon', 'Leonard', 'Penny', 'Rajesh', 'Howard']


# 일반적인 큐를 이용할때 메모리 부족 현상이 일어남
# 숫자가 같은 팀끼리 한팀으로 생각하면, 이름이 증가 횟수는 1, 2, 4, 8, 16....
# 즉 5, 10, 20, 40....끼리는 하나의 로테이션으로 볼 수 있음

# 예를 들어 45라는 숫자를 입력하면,
# 5+10+20 =35 < 45 < 5+10+20+40 =75 가 되기에
# 45 - 35(앞의 로테이션) = 10
# 즉 10이 로테이션이 끝난 뒤의 숫자이며, 4번째 로테이션(0, 1, 2, 3...)이기 때문에, 2^3만큼 같은 이름이 반복됨
# Sheldondl 8번 반복되고 2번 남은것은 Leonard이기 때문에 답은 Leonard

# 범위, 범위 합, 2의 지수
arrange = 5
s = 0
exp=0

while 1 :
	s = s+arrange
	arrange = arrange*2
	if s >= n : break
	exp=exp+1


# 나머지
remain = n-(s-arrange/2)

# 최종 횟수로 해당 위치의 사람 판단
point = ((remain-1)/pow(2,exp))%5

print name[point]




