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.util.*;
import java.io.*;
public class Solution {
	static FastScanner scr=new FastScanner();
	static PrintStream out=new PrintStream(System.out);
//	static ArrayList<Integer>list[];
	public static void main(String []args) {
//		int t=scr.nextInt();
//		while(t-->0) {
			solve();
//		}
//		out.println(sb);
	}
//	static boolean visited[][];
	static void solve() {
		long n=scr.nextLong();
		
		long sum=0;
		while(n>=19) {
			sum+=9+(n-9)%10;
			n=(n-9)/10;
		}
		sum+=n;
		out.println(sum);
	}
	static int gcd(int a,int b){
		if(b==0) {
			return a;
		}
		
		return gcd(b,a%b);
	}
	static long modPow(long base,long exp,long mod) {
		if(exp==0) {
			return 1;
		}
		if(exp%2==0) {
			long res=(modPow(base,exp/2,mod));
			return (res*res)%mod;
		}
		return (base*modPow(base,exp-1,mod))%mod;
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
	 	String next() {
	 		while (!st.hasMoreTokens())
	 			try {
	 				st=new StringTokenizer(br.readLine());
	 			} catch (IOException e) {
	 				e.printStackTrace();
	 			}
	 			return st.nextToken();
		}
		int[] sort(int a[]) {
			Arrays.sort(a);
			return a;
		}
		int []reverse(int a[]){
			int b[]=new int[a.length];
			int index=0;
			for(int i=a.length-1;i>=0;i--) {
				b[index]=a[i];
			}
			return b;
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long[] readLongArray(int n) {
			long  [] a=new long  [n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}

}