def have_3(s):
    for i in s:
        if(i==3):
            return 1
    return 0

d={}
for i,ma in enumerate([1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000,10000000000,100000000000,1000000000000,10000000000000,100000000000000,1000000000000000,10000000000000000,100000000000000000,1000000000000000000,10000000000000000000]):
    d[i]=ma

P=int(input())
k=d[P]
count=k//3
for i in range(2,k,3):
    if(have_3(list(map(int,list(str(i)))))==1):
        count+=1      
for i in range(4,k,3):
    if(have_3(list(map(int,list(str(i)))))==1):
        count+=1
print(count)