#include <bits/stdc++.h>

using namespace std;
using ll=long long;

int main()
{
   int n;cin >>n;
   int s=0;
   while(n>0){
    s+=n%10;
    n/=10;
   }
   if(s==1)cout << 10 << endl;
   else cout << s << endl;
    return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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".

Examples

Input

15


Output

6


Input

100000


Output

10

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;
using ll=long long;

int main()
{
   int n;cin >>n;
   int s=0;
   while(n>0){
    s+=n%10;
    n/=10;
   }
   if(s==1)cout << 10 << endl;
   else cout << s << endl;
    return 0;
}

```