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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.IO;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder.Contest.A
{
	static class Program
	{
		public static void Main(string[] args)
		{
			var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
			Console.SetOut(sw);

			Scanner cin = new Scanner();

			int n = cin.NextInt();
			int a = n / 20 * 10;
			int b = n - a;
			int ans = 0;
			while (a > 0)
			{
				ans += a % 10;
				a /= 10;
			}
			while (b > 0)
			{
				ans += b % 10;
				b /= 10;
			}
			Console.WriteLine(ans);

			Console.Out.Flush();
		}
	}

	class Scanner
	{
		string[] s;
		int i;

		char[] cs = new char[] { ' ' };

		public Scanner()
		{
			s = new string[0];
			i = 0;
		}

		public string Next()
		{
			if (i < s.Length) return s[i++];
			string st = Console.ReadLine();
			while (st == "") st = Console.ReadLine();
			s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
			if (s.Length == 0) return Next();
			i = 0;
			return s[i++];
		}

		public int NextInt()
		{
			return int.Parse(Next());
		}

		public int[] ArrayInt(int N, int add = 0)
		{
			int[] Array = new int[N];
			for (int i = 0; i < N; i++)
			{
				Array[i] = NextInt() + add;
			}
			return Array;
		}

		public long NextLong()
		{
			return long.Parse(Next());
		}

		public long[] ArrayLong(int N, long add = 0)
		{
			long[] Array = new long[N];
			for (int i = 0; i < N; i++)
			{
				Array[i] = NextLong() + add;
			}
			return Array;
		}

		public double NextDouble()
		{
			return double.Parse(Next());
		}

		public double[] ArrayDouble(int N, double add = 0)
		{
			double[] Array = new double[N];
			for (int i = 0; i < N; i++)
			{
				Array[i] = NextDouble() + add;
			}
			return Array;
		}
	}
}