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
"""
    Author: Sagar Pandey

"""
# ---------------------------------------------------Import Libraries---------------------------------------------------
import sys
import os
from math import sqrt, log, log2, log10, gcd, floor, pow, sin, cos, tan, pi, inf, factorial
from copy import copy, deepcopy
from sys import exit, stdin, stdout
from collections import Counter, defaultdict, deque
from itertools import permutations
import heapq
from bisect import bisect_left as bl
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect

# If the element is already present in the list,
# the right most position where element has to be inserted is r

# ---------------------------------------------------Global Variables---------------------------------------------------
# sys.setrecursionlimit(100000000)
mod = 1000000007
# ---------------------------------------------------Helper Functions---------------------------------------------------
iinp = lambda: int(sys.stdin.readline())
inp = lambda: sys.stdin.readline().strip()
strl = lambda: list(inp().strip().split(" "))
intl = lambda: list(map(int, inp().split(" ")))
mint = lambda: map(int, inp().split())
flol = lambda: list(map(float, inp().split(" ")))
flush = lambda: stdout.flush()


def permute(nums):
    def fun(arr, nums, cur, v):
        if len(cur) == len(nums):
            arr.append(cur.copy())
        i = 0
        while i < len(nums):
            if v[i]:
                i += 1
                continue
            else:
                cur.append(nums[i])
                v[i] = 1
                fun(arr, nums, cur, v)
                cur.pop()
                v[i] = 0
                i += 1
            # while i<len(nums) and nums[i]==nums[i-1]:i+=1    # Uncomment for unique permutations
        return arr

    res = []
    nums.sort()
    v = [0] * len(nums)
    return fun(res, nums, [], v)


def subsets(res, index, arr, cur):
    res.append(cur.copy())
    for i in range(index, len(arr)):
        cur.append(arr[i])
        subsets(res, i + 1, arr, cur)
        cur.pop()
    return res


def sieve(N):
    root = int(sqrt(N))
    primes = [1] * (N + 1)
    primes[0], primes[1] = 0, 0
    for i in range(2, root + 1):
        if primes[i]:
            for j in range(i * i, N + 1, i):
                primes[j] = 0
    return primes


def bs(arr, l, r, x):
    if x < arr[0] or x > arr[len(arr) - 1]:
        return -1
    while l <= r:
        mid = l + (r - l) // 2
        if arr[mid] == x:
            return mid
        elif arr[mid] < x:
            l = mid + 1
        else:
            r = mid - 1
    return -1


def isPrime(n):
    if n <= 1: return False
    if n <= 3: return True
    if n % 2 == 0 or n % 3 == 0: return False
    p = int(sqrt(n))
    for i in range(5, p + 1, 6):
        if n % i == 0 or n % (i + 2) == 0:
            return False
    return True


# -------------------------------------------------------Functions------------------------------------------------------

def solve():
    n=iinp()
    arr=["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
    m=1
    while 5*m<n:
        n-=5*m
        m*=2
    n-=1
    print(arr[n//m])



# -------------------------------------------------------Main Code------------------------------------------------------
solve()
