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>
#define REP(i, n) for(int i = 0;i < n;i++)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end())
#define VRSORT(v) sort(v.rbegin(), v.rend())//vectorの降順ソート
#define ll long long
#define pb(a) push_back(a)
#define INF 1000000000
#define MOD 1000000007
using namespace std;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;

typedef vector<unsigned int>vec;
typedef vector<vec> mat;
typedef tuple<ll, ll, ll> T;

int dy[]={0, 0, 1, -1, 0};
int dx[]={1, -1, 0, 0, 0};

int digitsum(int n){
    int ans=0;
    while(n>0){
        ans += n%10;
        n/=10;
    }
    return ans;
}
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    int ans=INF;
    for(int a=1;a<n;a++){
        int b=n-a;
        ans = min(ans,digitsum(a)+digitsum(b));
    }
    cout<<ans<<endl;

}