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.*;
import java.util.*;

public class B implements Runnable{
	public static void main (String[] args) {new Thread(null, new B(), "_cf", 1 << 28).start();}

	int[][] dp;
	long n;
	String ss;
	
	public void run() {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		System.err.println("Go!");

		n = fs.nextLong();
		ss = "" + n;
		dp = new int[ss.length()][111];
		for(int[] d : dp) Arrays.fill(d, -1);
		out.println(solve(0, 0));

		out.close();
	}
	
	int oo = (int)1e9;
	int solve(int digit, int carry) {
		if(digit == dp.length) {
			if(carry > 0) return -oo;
			return 0;
		}
		int res = -oo;
		if(dp[digit][carry] != -1) return dp[digit][carry];
		for(int i = 0; i <= 9; i++) {
			for(int j = 0; j <= 9; j++) {
				int gets = (i + j + carry) % 10;
				if(gets != ss.charAt(ss.length()-digit-1)-'0') continue;
				int ret = solve(digit+1, (i+j+carry) / 10);
				if(ret < 0) continue;
				res = Math.max(res, ret + i + j);
			}
		}
		return dp[digit][carry] =res; 
	}

	void sort (int[] a) {
		int n = a.length;
		for(int i = 0; i < 50; i++) {
			Random r = new Random();
			int x = r.nextInt(n), y = r.nextInt(n);
			int temp = a[x];
			a[x] = a[y];
			a[y] = temp;
		}
		Arrays.sort(a);
	}

	class FastScanner {
		public int BS = 1<<16;
		public char NC = (char)0;
		byte[] buf = new byte[BS];
		int bId = 0, size = 0;
		char c = NC;
		double num = 1;
		BufferedInputStream in;

		public FastScanner() {
			in = new BufferedInputStream(System.in, BS);
		}

		public FastScanner(String s) throws FileNotFoundException {
			in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
		}

		public char nextChar(){
			while(bId==size) {
				try {
					size = in.read(buf);
				}catch(Exception e) {
					return NC;
				}                
				if(size==-1)return NC;
				bId=0;
			}
			return (char)buf[bId++];
		}

		public int nextInt() {
			return (int)nextLong();
		}

		public long nextLong() {
			num=1;
			boolean neg = false;
			if(c==NC)c=nextChar();
			for(;(c<'0' || c>'9'); c = nextChar()) {
				if(c=='-')neg=true;
			}
			long res = 0;
			for(; c>='0' && c <='9'; c=nextChar()) {
				res = (res<<3)+(res<<1)+c-'0';
				num*=10;
			}
			return neg?-res:res;
		}

		public double nextDouble() {
			double cur = nextLong();
			return c!='.' ? cur:cur+nextLong()/num;
		}

		public String next() {
			StringBuilder res = new StringBuilder();
			while(c<=32)c=nextChar();
			while(c>32) {
				res.append(c);
				c=nextChar();
			}
			return res.toString();
		}

		public String nextLine() {
			StringBuilder res = new StringBuilder();
			while(c<=32)c=nextChar();
			while(c!='\n') {
				res.append(c);
				c=nextChar();
			}
			return res.toString();
		}

		public boolean hasNext() {
			if(c>32)return true;
			while(true) {
				c=nextChar();
				if(c==NC)return false;
				else if(c>32)return true;
			}
		}
		
		public int[] nextIntArray(int n) {
			int[] res = new int[n];
			for(int i = 0; i < n; i++) res[i] = nextInt();
			return res;
		}
		
	}

}