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 <algorithm>
#include <iostream>
#include <vector>
#include <math.h>
#include <set>
#include <map>
#include <string>
#include <stack>
#include <queue>
#include <iomanip>
#include <numeric>
#include <tuple>
#define _USE_MATH_DEFINES
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> plglg;
typedef tuple<int, int, int> tiii;
typedef tuple<ll, ll, ll> tlglglg;
double pi = 3.141592653589793;
ll mod = 1000000007;
int intmax = 2147483647;
int intmin = -2147483648;
ll llmax = 9223372036854775807;
ll llmin = -9223372036854775807;

int main() {
    ll N;
    cin >> N;
    ll ans = 100000;
    for (ll i = 1; i < N; i++) {
        ll p = i;
        ll j = N - i;
        ll inum = 0;
        ll jnum = 0;
        for (int k = 0; k < 7; k++) {
            inum += p % 10;
            jnum += j % 10;
            p = p / 10;
            j = j / 10;
        }
        if (inum + jnum < ans) {
            ans = inum + jnum;
        }
    }
    cout << ans << endl;
}