#思路:把1000以内的素数算出来存到表里,再查表
import math
def check(k):
    for i in range(2,int(math.sqrt(k))+1):
        if k%i==0:
            return False
    return True
l=[2]
for i in range(3,1001):
    if check(i):
        l.append(i)
n=int(input())
ans1=ans2=0
min=10000
for i in range(len(l)):
    for j in range(i,len(l)):
        if l[i]+l[j]==n and l[j]-l[i]<min:
            ans1=l[i]
            ans2=l[j]
            min=l[j]-l[i]
print(ans1)
print(ans2)