Given is a positive integer N. How many tuples (A,B,C) of positive integers satisfy A \times B + C = N?

Constraints

* 2 \leq N \leq 10^6
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N


Output

Print the answer.

Examples

Input

3


Output

3


Input

100


Output

473


Input

1000000


Output

13969985
MAX = 1000001; 
  
# array to store 
# prime factors 
factor = [0]*(MAX + 1); 
  
# function to generate all  
# prime factors of numbers 
# from 1 to 10^6 
def generatePrimeFactors(): 
    factor[1] = 1; 
  
    # Initializes all the  
    # positions with their value. 
    for i in range(2,MAX): 
        factor[i] = i; 
  
    # Initializes all  
    # multiples of 2 with 2 
    for i in range(4,MAX,2): 
        factor[i] = 2; 
  
    # A modified version of  
    # Sieve of Eratosthenes 
    # to store the smallest  
    # prime factor that divides 
    # every number. 
    i = 3; 
    while(i * i < MAX): 
        # check if it has 
        # no prime factor. 
        if (factor[i] == i): 
            # Initializes of j  
            # starting from i*i 
            j = i * i; 
            while(j < MAX):  
                # if it has no prime factor  
                # before, then stores the  
                # smallest prime divisor 
                if (factor[j] == j): 
                    factor[j] = i; 
                j += i; 
        i+=1; 
  
# function to calculate 
# number of factors 
def calc(n): 
    if (n == 1): 
        return 1; 
    ans = 1; 
  
    # stores the smallest  
    # prime number that 
    # divides n 
    dup = factor[n]; 
  
    # stores the count of  
    # number of times a  
    # prime number divides n. 
    c = 1; 
  
    # reduces to the next  
    # number after prime 
    # factorization of n 
    j = int(n / factor[n]); 
  
    # false when prime  
    # factorization is done 
    while (j > 1): 
        # if the same prime  
        # number is dividing  
        # n, then we increase 
        # the count 
        if (factor[j] == dup): 
            c += 1; 
  
        # if its a new prime factor 
        # that is factorizing n,  
        # then we again set c=1 and  
        # change dup to the new prime  
        # factor, and apply the formula  
        # explained above.  
        else: 
            dup = factor[j]; 
            ans = ans * (c + 1); 
            c = 1; 
  
        # prime factorizes 
        # a number 
        j = int(j / factor[j]); 
  
    # for the last 
    # prime factor 
    ans = ans * (c + 1); 
  
    return ans;
generatePrimeFactors()
n=input()
c=0
for i in range(1,n):
    c+=calc(n-i)
print c
