#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)