n = int(input())
count = 0
q = 5 # 从5^1开始处理
while q <= n:
max_a = n // q # 当前q的最大指数
# 计算sum_{k=1}^n floor(k/q)的总和
count += (max_a * (max_a - 1) // 2) * q + max_a * (n - max_a * q + 1)
q *= 5 # 处理下一个5的幂次,如5^2=25,5^3=125等
print(count)
biA9IGludChpbnB1dCgpKQoKY291bnQgPSAwCnEgPSA1ICAjIOS7jjVeMeW8gOWni+WkhOeQhgoKd2hpbGUgcSA8PSBuOgogICAgbWF4X2EgPSBuIC8vIHEgICMg5b2T5YmNceeahOacgOWkp+aMh+aVsAogICAgIyDorqHnrpdzdW1fe2s9MX1ebiBmbG9vcihrL3Ep55qE5oC75ZKMCiAgICBjb3VudCArPSAobWF4X2EgKiAobWF4X2EgLSAxKSAvLyAyKSAqIHEgKyBtYXhfYSAqIChuIC0gbWF4X2EgKiBxICsgMSkKICAgIHEgKj0gNSAgIyDlpITnkIbkuIvkuIDkuKo155qE5bmC5qyh77yM5aaCNV4yPTI177yMNV4zPTEyNeetiQoKcHJpbnQoY291bnQp