'''
dict的复杂度大都在O(1)
'''
import math
import sys
n = int(input())

mx = int(math.sqrt(n)) + 2

key = {}
for c in range(mx):
    for d in range(mx):
        key[c**2 + d**2] = c
for a in range(mx):
    for b in range(mx):
        left = n - a**2 - b**2
        c = key.get(left, -1)
        if c != -1:
            d = int(math.sqrt(left - c ** 2))
            a, b, c, d = sorted([a, b, c, d])
            print(a, b, c, d)
            sys.exit()