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
# Lets goto the next level 
# AIM Specialist at CF *__* asap 
# template taken from chaudhary_19  
# Remember you were also a novice when you started, 
# hence never be rude to anyone who wants to learn something
# Never open a ranklist untill and unless you are done with solving problems, wastes 3/4 minuts 
# Donot treat CP as a placement thing, love it and enjoy it, you will succeed for sure.  
# Any doubts or want to have a talk, contact https://www.facebook.com/chaudhary.mayank


# ///==========Libraries, Constants and Functions=============///

import sys
from bisect import bisect_left,bisect_right,insort
from collections import deque,Counter
from math import gcd,sqrt,factorial,ceil,log10,log2,floor
from itertools import permutations
from heapq import heappush,heappop,heapify
inf = float("inf")
mod = 1000000007
#sys.setrecursionlimit(10**5)


def factorial_p(n, p):
    ans = 1
    if n <= p // 2:
        for i in range(1, n + 1):
            ans = (ans * i) % p
    else:
        for i in range(1, p - n):
            ans = (ans * i) % p
        ans = pow(ans, p - 2, p)
        if n % 2 == 0:
            ans = p - ans
    return ans


def nCr_p(n, r, p):
    ans = 1
    while (n != 0) or (r != 0):
        a, b = n % p, r % p
        if a < b:
            return 0
        ans = (ans * factorial_p(a, p) * pow(factorial_p(b, p), p - 2, p) * pow(factorial_p(a - b, p), p - 2, p)) % p
        n //= p
        r //= p
    return ans

def prime_sieve(n):
    """returns a sieve of primes >= 5 and < n"""
    flag = n % 6 == 2
    sieve = bytearray((n // 3 + flag >> 3) + 1)
    for i in range(1, int(n**0.5) // 3 + 1):
        if not (sieve[i >> 3] >> (i & 7)) & 1:
            k = (3 * i + 1) | 1
            for j in range(k * k // 3, n // 3 + flag, 2 * k):
                sieve[j >> 3] |= 1 << (j & 7)
            for j in range(k * (k - 2 * (i & 1) + 4) // 3, n // 3 + flag, 2 * k):
                sieve[j >> 3] |= 1 << (j & 7)
    return sieve
def prime_list(n):
    """returns a list of primes <= n"""
    res = []
    if n > 1:
        res.append(2)
    if n > 2:
        res.append(3)
    if n > 4:
        sieve = prime_sieve(n + 1)
        res.extend(3 * i + 1 | 1 for i in range(1, (n + 1) // 3 + (n % 6 == 1)) if not (sieve[i >> 3] >> (i & 7)) & 1)
    return res
    
        
def binary(number):                                             #  <----- calculate the no. of 1's in binary representation of number
    result=0
    while number:
        result=result+1
        number=number&(number-1)
    return result

def is_prime(n):
    """returns True if n is prime else False"""
    if n < 5 or n & 1 == 0 or n % 3 == 0:
        return 2 <= n <= 3
    s = ((n - 1) & (1 - n)).bit_length() - 1
    d = n >> s
    for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
        p = pow(a, d, n)
        if p == 1 or p == n - 1 or a % n == 0:
            continue
        for _ in range(s):
            p = (p * p) % n
            if p == n - 1:
                break
        else:
            return False
    return True

def all_factors(n):
    """returns a sorted list of all distinct factors of n"""
    small, large = [], []
    for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
        if not n % i:
            small.append(i)
            large.append(n // i)
    if small[-1] == large[-1]:
        large.pop()
    large.reverse()
    small.extend(large)
    return small

def expo(x,n):       # x ki power n
    if n==0:
        return 1
    result=1
    while n>1:
        if n&1:
            result=result*x
            n-=1
        else:
            n//=2
            x=x*x
    return (result*x)

def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
# ///===========MAIN=============///
mydict={1:'Sheldon',2:'Leonard',3:'Penny',4:'Rajesh',5:'Howard'}
n=int(input())
power=0;total=0
while ((total+(5*expo(2,power)))<=n):
    total+=(5*expo(2,power))
    power+=1
if n==total:
    print("Howard")
    exit()
nxt=total+(5*expo(2,power))
z=int(ceil((n-total)/((nxt-total)//5)))
print(mydict[z])
