# a, b で全探索
# a, bが決まればそのときのp進数がどこまで行けるかも決まるはず
# p進数分数は、[a,b] (in p) / [p-1,p-1] (in p)と考えればいい
# https://detail.chiebukuro.yahoo.co.jp/qa/question_detail/q12186530002
# 二分探索で高速化

# https://science-log.com/python%E3%83%97%E3%83%AD%E3%82%B0%E3%83%A9%E3%83%9F%E3%83%B3%E3%82%B0tips%E9%9B%86/%E3%80%90python%E3%80%91%E8%A8%98%E6%95%B0%E6%B3%95%E3%81%AE%E5%A4%89%E6%8F%9B/
def N_to_Dec(digits, base):
    num = 0
    for digit in digits:
        num = num * base + digit
    return num

N = int(input())
ans = 0
for a in range(0, 10):
    for b in range(0, 10):
        if a == b:
            continue 
        
        OK = 1
        NG = 10**20
        while NG-OK > 1:
            mid = (NG+OK)//2
            numerator = N_to_Dec([a, b], mid)
            denominator = N_to_Dec([mid-1, mid-1], mid)
            if numerator*N > denominator:
                OK = mid
            else:
                NG = mid  
        #print('a', a, 'b', b, 'OK', OK)
        ans += max(0, OK - max(a, b))
print(ans)