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)

#include<bits/stdc++.h>
using namespace std;

void doRoutine() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    srand(322179);
#ifdef LOCAL
    freopen("/root/alg/clion_main/input.txt", "rt", stdin);
    freopen("/root/alg/clion_main/output.txt", "wt", stdout);
#endif
}

#define DBG(VAR) cerr << #VAR << ": " << VAR << endl

#define all(x) x.begin(), x.end()
#define last(x) (--x.end())
#define vb vector<bool>
#define vi vector<int>
#define vvi vector<vi>
#define pii pair<int, int>
#define fi first
#define se second
#define pb push_back
#define vc vector

#ifndef LOCAL
#define cerr if(0) cerr
#define DBG(...) (true)
#endif

#define D double
#define int long long

// ----------------------------- CODE NOW ------------------------


const int INF = 1000000000000000000;

int sum(int x) {
    int res = 0;
    while (x > 0) {
        res += x % 10;
        x /= 10;
    }
    return res;
}

signed main() {
    doRoutine();

    int n;
    cin >> n;
    int ans = INF;
    for (int a = 1; a < n; ++a) {
        ans = min(ans, sum(a) + sum(n - a));
    }

    cout << ans << endl;

    return 0;
}