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.*;
import java.util.*;

public class Main {
    public static final long mod = (long)1e9+7;
    public static final long INF = Long.MAX_VALUE/10;
    public static final int inf = Integer.MAX_VALUE/10;

    static void solve(InputReader in, PrintWriter out){
        int n = in.ni();
        int min = inf;
        int sum = 0;
        for (int i = 1; i < n; i++) {
            String a = String.valueOf(i);
            String b = String.valueOf(n-i);
            sum = 0;
            for (int j = 0; j < a.length(); j++) {
                sum += a.charAt(j)-'0';
            }
            for (int j = 0; j < b.length(); j++) {
                sum += b.charAt(j)-'0';
            }
            min = Math.min(min, sum);
        }
        out.println(min);
    }

    public static void main(String[] args) throws Exception{
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        solve(in, out);
        out.close();
    }
    public static class InputReader{
        private BufferedReader br;
        private StringTokenizer st;
        public InputReader(InputStream is){
            br = new BufferedReader(new InputStreamReader(is));
            st = null;
        }
        public String ns(){
            if(st == null || !st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (Exception e){
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }
        public long nl(){
            return Long.parseLong(ns());
        }
        public int ni(){
            return Integer.parseInt(ns());
        }
        public Double nd(){
            return Double.parseDouble(ns());
        }
        public int[] ni(int n){
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }
        public long[] nl(int n){
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nl();
            }
            return a;
        }
        public double[] nd(int n){
            double[] a = new double[n];
            for (int i = 0; i < n; i++) {
                a[i] = nd();
            }
            return a;
        }
    }
}