#include <iostream>

#define REP(i,n) for(int i=0; i<n; i++)

using namespace std;

int main(){
	string n;
	int ans;
	
	ans = 0;
	cin >> n;
	REP(i,n.length()){
		ans += n[i]-'0';
	}
	if (ans<=1) ans = 10;
	cout << ans << endl;
	return 0;
}

### Prompt

In Cpp, your task is to solve 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 <iostream>

#define REP(i,n) for(int i=0; i<n; i++)

using namespace std;

int main(){
	string n;
	int ans;
	
	ans = 0;
	cin >> n;
	REP(i,n.length()){
		ans += n[i]-'0';
	}
	if (ans<=1) ans = 10;
	cout << ans << endl;
	return 0;
}

```