calculate count of numbers mod 3

# algoritm of calculate count of numbers in
# row like 1, 12, 123, 1234, .. 123456789, 12345678910, 1234567891011.. N
# thats mod 3

import time

i = int(input('input N: '))

number   = 0
counter  = 0

t1 = time.monotonic()
for i in range(1, i+1):
    # detect rank of iteration number
    rank  = 1
    buf_i = i
    while True:
        if buf_i // 10:
            rank += 1
            buf_i = buf_i // 10
        else:
            break
    number = number*(10**rank) + i
    # calculate sum of digits and check it mod 3
    x = number
    x = (x >> 16) + (x & 0xffff)
    x = (x >> 10) + (x & 0x3ff)
    x = (x >> 6)  + (x & 0x3f)
    x = (x >> 4)  + (x & 0xf)
    while x >3:
        x = (x >> 2) + (x & 0x3)
    if (x == 3):
        counter += 1
    # old style
    '''
    num  = number
    summ = 0
    while num:
        num, mod = divmod(num, 10)
        summ += mod
    if summ % 3 == 0:
        counter += 1
    '''

t2 = time.monotonic()
delta = t2-t1
print('%.6fs' % delta, counter)

# input N: 1000
# 1.653000s 666