Problem:
Score : 200 points

Problem Statement

Takahashi has two positive integers A and B.

It is known that A plus B equals N.
Find the minimum possible value of "the sum of the digits of A" plus "the sum of the digits of B" (in base 10).

Constraints

2 ≤ N ≤ 10^5

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the minimum possible value of "the sum of the digits of A" plus "the sum of the digits of B".

Sample Input 1

15

Sample Output 1

6

When A=2 and B=13, the sums of their digits are 2 and 4, which minimizes the value in question.

Sample Input 2

100000

Sample Output 2

10

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _USE_MATH_DEFINES
#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <string>
#include <sstream>
#include <iomanip>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <bitset>
#include <utility>
#include <numeric>
#include <iterator>

using namespace std;

using ll = long long;
using ull = unsigned long long;

const ll inf = 1ll << 60;
const ll mod = (ll)1e9 + 7;

#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()

#define print(s) cout << s;
#define println(s) cout << s << endl;
#define printd(s, f) cout << fixed << setprecision(f) << s << endl;


int main()
{
	ll n;
	cin >> n;

	ll res = inf;
	for (ll i = 1; i <= n / 2; i++)
	{
		ll j = n - i;

		ll s = 0;
		for (ll p = i; p > 0; p /= 10)
		{
			s += p % 10;
		}

		for (ll p = j; p > 0; p /= 10)
		{
			s += p % 10;
		}

		res = min(res, s);
	}
	println(res);
}