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

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
	private static final int MOD = (int)Math.pow(10, 9);
	public static void main(String[] args) {
		FastReader sc = new FastReader();
		int n = sc.nextInt();
		int base = 1;
		int res = Integer.MAX_VALUE;
		
		while (base < n) {
			int remain = n - base;
			res = Math.min(res, findDigit(remain) + 1);
			base *= 10;
		}
		System.out.println(res);
	}
	
	private static int findDigit(int num) {
		int res = 0;
		while (num > 0) {
			res += (num % 10);
			num /= 10;
		}
		return res;
	}
	
	static class FastReader {
	    BufferedReader br;
	    StringTokenizer st;
	
	    public FastReader() {
	      br = new BufferedReader(new InputStreamReader(System.in));
	    }
	  
	    String next() { 
	      while (st == null || !st.hasMoreElements()) {
	        try {
	          st = new StringTokenizer(br.readLine());
	        } catch (IOException  e) {
	          e.printStackTrace();
	        }
	      }
	      return st.nextToken();
	    }
	
	    int nextInt() {
	      return Integer.parseInt(next());
	    }
	
	    long nextLong() {
	      return Long.parseLong(next());
	    }
	
	    double nextDouble() { 
	      return Double.parseDouble(next());
	    }
	
	    String nextLine() {
	      String str = "";
	        try{
	            str = br.readLine();
	        } catch (IOException e) {
	            e.printStackTrace();
	        }
	        return str;
	      }
	  }
}
