Problem: Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?

Please help Haruhi solve this problem.

Input

The first line of input will be a single string s (1 ≤ |s| ≤ 20). String s consists only of lowercase English letters. 

Output

Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.

Examples

Input

a


Output

51


Input

hi


Output

76

Note

In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets. 

Solution (in JAVA):

Solution:
import java.io.BufferedReader;

import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.*;
import static java.lang.Math.sqrt;
import static java.lang.Math.round;
import java.lang.Math;
public class Jo {
	public static void main(String args[]) throws IOException 
    { 
		Scanner sc=new Scanner(System.in);
		String s=sc.next();
	    int c=((s.length()+1)*25)+1;
	    System.out.println(c);	
    }
	//count number of digits of number n
	static long countDigit(long n) 
    { 
        return (long)Math.floor(Math.log10(n) + 1); 
    } 
    //min. summation of pairs of divisors
	public static int di(int n) {
		int s;int min=1000000;
		for(int i=1;i*i<=n;i++) {
			if(n%i==0) {
				s=i+(n/i);
				if(s<=min)min=s;
			}
		}
		return min;
	}
	
	public static double squareRoot(int number) {
		double temp;

		double sr = number / 2;

		do {
			temp = sr;
			sr = (temp + (number / temp)) / 2;
		} while ((temp - sr) != 0);

		return sr;
	    }
	static boolean isPerfect(int n) 
    { 
        // To store sum of divisors 
        int sum = 1; 
      
        // Find all divisors and add them 
        for (int i = 2; i * i <= n; i++) 
        { 
            if (n % i==0) 
            { 
                if(i * i != n) 
                    sum = sum + i + n / i; 
                else
                    sum = sum + i; 
            } 
        }  
        // If sum of divisors is equal to 
        // n, then n is a perfect number 
        if (sum == n && n != 1) 
            return true; 
      
        return false; 
    } 
}