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
# WHERE: https://codeforces.com/problemset/page/8?order=BY_RATING_ASC

# Taxi
# https://codeforces.com/problemset/problem/158/B
# Input
# The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
# Output
# Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
# ignoreInput = input()
# groups = list(map(int, input().split()))
# i = 0
# j = len(groups) - 1  # last position of the array
# counter = 0

# counters = [0, 0, 0, 0]

# for group in groups:
#     counters[(group-1)] += 1

# fours = counters[3]
# threes = counters[2]
# ones = 0
# if counters[0] > threes:
#     ones = counters[0] - threes
# twos = (counters[1]//2) + (((counters[1] % 2)*2+ones)//4)
# # left = 0
# if (((counters[1] % 2)*2+ones) % 4) != 0:
#     twos += 1

# print(fours+threes+twos)
# ---------------------------------------------------------------------------------------------------------------------

# Fancy Fence
# https://codeforces.com/problemset/problem/270/A
# Input
# The first line of input contains an integer t (0 < t < 180) — the number of tests. Each of the following t lines contains a single integer a (0 < a < 180) — the angle the robot can make corners at measured in degrees.
# Output
# For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.

# times = int(input())
# answers = []
# for time in range(times):
#     a = int(input())
#     if 360 % (180 - a) == 0:
#         answers.append("YES")
#     else:
#         answers.append("NO")
# for answer in answers:
#     print(answer)
# ---------------------------------------------------------------------------------------------------------------------

# Interesting drink
# https://codeforces.com/problemset/problem/706/B
# Input
# The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of shops in the city that sell Vasiliy's favourite drink.
# The second line contains n integers xi (1 ≤ xi ≤ 100 000) — prices of the bottles of the drink in the i-th shop.
# The third line contains a single integer q (1 ≤ q ≤ 100 000) — the number of days Vasiliy plans to buy the drink.
# Then follow q lines each containing one integer mi (1 ≤ mi ≤ 109) — the number of coins Vasiliy can spent on the i-th day.
# Output
# Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.

# nShops = int(input())
# shopPrices = list(map(int, input().split()))
# shopPrices.sort()
# times = int(input())
# answers = []


# def binarySearch(array, target, carry):
#     index = len(array)
#     if len(array) == 0:
#         return carry
#     if index == 1:
#         if target < array[0]:
#             return carry
#         else:
#             return carry + 1
#     # position in the middle
#     index = index//2
#     if target < array[index]:
#         # return the left
#         newPrices = array[0:index]
#         return binarySearch(newPrices, target, carry)
#     else:
#         # return the right
#         carry += (index)
#         newPrices = array[index:]
#         return binarySearch(newPrices, target, carry)


# def iterativeBinary(array, target):
#     low = 0
#     high = len(array) - 1
#     while (low <= high):
#         mid = low + ((high-low)//2)
#         if array[mid] > target:
#             high = mid - 1
#         else:
#             low = mid + 1

#     return low


# for time in range(times):
#     money = int(input())
#     # looks like the way i implemented the binary search isnt logN :(
#     # buys = binarySearch(shopPrices, money, 0)
#     buys = iterativeBinary(shopPrices, money)
#     answers.append(buys)
# for answer in answers:
#     print(answer)
# ---------------------------------------------------------------------------------------------------------------------

# A and B and Compilation Errors
# https://codeforces.com/problemset/problem/519/B
# Output
# Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.

# compilationTimes = int(input())
# errors1 = list(map(int, input().split()))
# errors2 = list(map(int, input().split()))
# errors3 = list(map(int, input().split()))

# errors1.sort()
# errors2.sort()
# errors3.sort()

# n = -1
# isNFound = False
# m = 1
# isMFound = False
# for time in range(compilationTimes):
#     if isNFound and isMFound:
#         break
#     if not isNFound:
#         if time < len(errors2):
#             if errors1[time] != errors2[time]:
#                 n = errors1[time]
#                 isNFound = True
#         else:
#             n = errors1[time]
#             isNFound = True

#     if not isMFound:
#         if time < len(errors3):
#             if errors2[time] != errors3[time]:
#                 m = errors2[time]
#                 isMFound = True
#         else:
#             m = errors2[time]
#             isNFound = True


# print(n)
# print(m)
# ---------------------------------------------------------------------------------------------------------------------


target = int(input())
people = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
target -= 1
k = 0
while True:
    if 5 * (2**k - 1) > target:
        break
    k += 1
k -= 1
# print(k)
# print((5 * (2**k - 1)))
corrimiento = target - (5 * (2**k - 1))
realK = 2**(k)
positionArray = (corrimiento)//realK
# print(realK)
# print(corrimiento)
print(people[positionArray])


# ---------------------------------------------------------------------------------------------------------------------
