'''
解题思路:
构造一个素数判断函数isSu(n),搜索范围为 2 -> int(n**0.5+1)
两个数i、j相加搜索,搜索范围取一半 0 -> n//2+1
'''
def isSu(n):
for i in range(2,int(n**0.5+1)):
if n%i==0:
return False
return True
while 1:
try:
pass
n = int(input())
#print('n=',n)
dis = n
for i in range(n//2+1):
#print(i,isSu(i))
j = n-i
if isSu(i) and isSu(j) and abs(i-j)<dis:
i_min = i
j_min = j
print(i_min)
print(j_min)
except:
break

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