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:
def digit_sum(x):
    ans = 0
    while x > 0:
        ans += x%10
        x = x//10
    return ans

def digit_count(x):
    ans = 0
    while x > 0:
        ans += 1
        x = x//10
    return ans


n = int(input())
a, b = 0, 0
ans = 0
##for i in range(n+1):
##    print(i, n-i, digit_sum(i) + digit_sum(n - i))
##    ans = max(ans, digit_sum(i) + digit_sum(n - i))
##print(ans)
##if n % 2 == 0:
##    a = n//2 + 1
##    b = n//2 - 1    
##else:
##    a = n//2
##    b = n//2 + 1
##while a%10 != 9 and a >= 10:
##    a -= 1
##    b += 1
##    if (a == 0):
##        break
if n <= 18:
    print(n)
else:
    new_num = 1
    for i in range(digit_count(n)-1):
        new_num = new_num *10 + 9
    if (new_num > n):
        new_num = 9
        for i in range(digit_count(n) - 2):
            new_num = new_num * 10 + 9
##    print(new_num, n - new_num)
    print(digit_sum(new_num) + digit_sum(n - new_num))
##print(digit_sum(a) + digit_sum(b))