Problem: You are given a positive integer n.

Let S(x) be sum of digits in base 10 representation of x, for example, S(123) = 1 + 2 + 3 = 6, S(0) = 0.

Your task is to find two integers a, b, such that 0 ≤ a, b ≤ n, a + b = n and S(a) + S(b) is the largest possible among all such pairs.

Input

The only line of input contains an integer n (1 ≤ n ≤ 10^{12}).

Output

Print largest S(a) + S(b) among all pairs of integers a, b, such that 0 ≤ a, b ≤ n and a + b = n.

Examples

Input

35


Output

17


Input

10000000000


Output

91

Note

In the first example, you can choose, for example, a = 17 and b = 18, so that S(17) + S(18) = 1 + 7 + 1 + 8 = 17. It can be shown that it is impossible to get a larger answer.

In the second test example, you can choose, for example, a = 5000000001 and b = 4999999999, with S(5000000001) + S(4999999999) = 91. It can be shown that it is impossible to get a larger answer.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;


public class SumOFDIGITCORRECT {

	/**
	 * @param args
	 * @throws IOException 
	 */
	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		long num=Long.parseLong(br.readLine());
		ArrayList<Long> rs=new ArrayList<Long>();
		long h=num/2;
		String z="1";
		String th="1";
		char ch=Long.toString(num).charAt(0);
		String ss=Long.toString(Long.parseLong(ch+"")-1);
//		for(int i=1;i<Long.toString(h).length();i++){
//			z+="0";
//		}
		for(long i=1;i<Long.toString(num).length();i++){
			ss+="9";
		}
		
		long thl=Long.parseLong(ss);
		String smn="",sthl="";
		if(thl<num&& thl!=1){
			long mn=num-thl;
			smn=Long.toString(mn);
			sthl=Long.toString(thl);
			long r=0;
			for(int j=0;j<smn.length();j++){
				r+=Long.parseLong(smn.substring(j,j+1));
				//System.out.println(in.substring(j,j+1));
			}
			for(int j=0;j<sthl.length();j++){
				r+=Long.parseLong(sthl.substring(j,j+1));
				//System.out.println(in.substring(j,j+1));
			}
			
			//System.out.println("If");
			rs.add(r);
			thl+=Long.parseLong(z);
		}
//		else {
//			if(rs.size()==0){
//				if(num%2==0){
//				smn=Long.toString(h+1);
//				sthl=Long.toString(h-1);}
//				else{
//					smn=Long.toString(h);
//					sthl=Long.toString(h+1);
//				}
//				long r=0;
//				for(int j=0;j<smn.length();j++){
//					r+=Long.parseLong(smn.substring(j,j+1));
//					//System.out.println(in.substring(j,j+1));
//				}
//				for(int j=0;j<sthl.length();j++){
//					r+=Long.parseLong(sthl.substring(j,j+1));
//					//System.out.println(in.substring(j,j+1));
//				}
//				
//				//System.out.println("else");
//				rs.add(r);
//			}
			
		//}
		Collections.sort(rs);
		System.out.println(rs.get(rs.size()-1));
		
	}

}