Problem: You are given a positive integer n.

Let S(x) be sum of digits in base 10 representation of x, for example, S(123) = 1 + 2 + 3 = 6, S(0) = 0.

Your task is to find two integers a, b, such that 0 ≤ a, b ≤ n, a + b = n and S(a) + S(b) is the largest possible among all such pairs.

Input

The only line of input contains an integer n (1 ≤ n ≤ 10^{12}).

Output

Print largest S(a) + S(b) among all pairs of integers a, b, such that 0 ≤ a, b ≤ n and a + b = n.

Examples

Input

35


Output

17


Input

10000000000


Output

91

Note

In the first example, you can choose, for example, a = 17 and b = 18, so that S(17) + S(18) = 1 + 7 + 1 + 8 = 17. It can be shown that it is impossible to get a larger answer.

In the second test example, you can choose, for example, a = 5000000001 and b = 4999999999, with S(5000000001) + S(4999999999) = 91. It can be shown that it is impossible to get a larger answer.

Solution (in PYTHON3):

Solution:
from math import log10

n = int(input())

def sumsum(i, j):
    ii = list(map(int, list(str(i))))
    jj = list(map(int, list(str(j))))
    return(sum(ii) + sum(jj))

def brute_force(n):
    max_sum = 0
    for i in range(n + 1):
        j = n - i
        cur_sum = sumsum(i, j)
        if cur_sum > max_sum:
            max_sum = cur_sum
    return max_sum

def for_clever(n):
    if n == 0:
        return 0
    n_order = int(log10(n))
    cand1 = (10 ** n_order) - 1
    cand2 = n - cand1
    return sumsum(cand1, cand2)

#for n in range(10000):
#    if for_clever(n) != brute_force(n):
#        print(n, for_clever(n), brute_force(n))
#    if n % 1000 == 0:
#        print(".")

print(for_clever(n))