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.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
import java.lang.*;
import java.util.ArrayList;
import java.math.BigInteger;
public class me {

public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner br = new Scanner(System.in);
        String s=br.next();
        System.out.print((s.length() + 1) * 25 + 1);
}
     
     

    public static long  ways(int n, int r, long MOD)
    {
      //  vector<long long> f(n + 1,1);
        long f[]=new long[n+1];
        Arrays.fill(f,1);
        for (int i=2; i<=n;i++)
            f[i]= (f[i-1]*i) % MOD;
        return (f[n]*((InverseEuler(f[r], MOD) * InverseEuler(f[n-r], MOD)) % MOD)) % MOD;
    }  
    
                  
                  
                  
                   /*THIS IS FOR THE INVERSE EULER*/
    public static long  InverseEuler(long n, long MOD)
    {
        return pow(n,MOD-2,MOD);
    }
    
    
    
                   
                   /*THIS ID THE CHECK FOR THE PRIME*/
    static int isPrime(int n) {
    for(int i=2;i<n;i++) {
        if(n%i==0)
            return 0;
    }
    return 1;
}
    
    
    
                          /*TO CHECK FOR THE GCD IN LONG*/
public static long gcd(long x,long y)
{
    if(x%y==0)
        return y;
    else
        return gcd(y,x%y);
}
    
    
    
                             /* CHECK FOR THE GCD IN INT*/
public static int gcd(int x,int y)
{
    if(x%y==0)
        return y;
    else 
        return gcd(y,x%y);
}


                         /*CHECK FOR THE ABSOLUTE IN INT*/
public static int abs(int a,int b)
{
    return (int)Math.abs(a-b);
}



                        /*CHECK FOR THE ABSOLUTE IN LONG*/
public static long abs(long a,long b)
{
    return (long)Math.abs(a-b);
}


                       /*CHECK FOR MAX BETWEEN TWO NUMBERS IN INT*/
public static int max(int a,int b)
{
    if(a>b)
        return a;
    else
        return b;
}



                       /*CHECK FOR MIN BETWEEN TWO NUMBERS IN INT*/
public static int min(int a,int b)
{
    if(a>b)
        return b;
    else 
        return a;
}


                       /*CHECK FOR MAX BETWEEN TWO NUMBERS IN LONG*/
public static long max(long a,long b)
{
    if(a>b)
        return a;
    else
        return b;
}

                     /*CHECK FOR MIN BETWEEN TWO NUMBERS IN LONG*/
public static long min(long a,long b)
{
    if(a>b)
        return b;
    else 
        return a;
}
                    /*check for power of two of any num*/
static boolean isPowerOfTwo (long v) {
    return (v & (v - 1)) == 0;
}
public static long pow(long n,long p,long m)
{
     long  result = 1;
      if(p==0)
        return 1;
    if (p==1)
        return n;
    while(p!=0)
    {
        if(p%2==1)
            result *= n;
        if(result>=m)
        result%=m;
        p >>=1;
        n*=n;
        if(n>=m)
        n%=m;
    }
    return result;
}

                    /*SORT GIVEN ARRAY*/
static long sort(int a[],int n)
{   int b[]=new int[n]; 
    return mergeSort(a,b,0,n-1);}


static long mergeSort(int a[],int b[],long left,long right)
{ long c=0;if(left<right)
 {   long mid=left+(right-left)/2;
     c= mergeSort(a,b,left,mid);
     c+=mergeSort(a,b,mid+1,right);
     c+=merge(a,b,left,mid+1,right); }  
    return c;    }


static long merge(int a[],int  b[],long left,long mid,long right)
{long c=0;int i=(int)left;int j=(int)mid; int k=(int)left;
while(i<=(int)mid-1&&j<=(int)right)
{ if(a[i]<=a[j]) {b[k++]=a[i++]; }
else    { b[k++]=a[j++];c+=mid-i;}}
while (i <= (int)mid - 1)   b[k++] = a[i++]; 
while (j <= (int)right) b[k++] = a[j++];
for (i=(int)left; i <= (int)right; i++) 
    a[i] = b[i];  return c;  }


    
    
                       /*REMOVE ZERO FROM GIVEN NUM*/
static int removeZero(int num) {
        int ret = 0;
        int ten = 1;
        while (num>0) {
                int dig = num % 10;
                num /= 10;
                if (dig>0) {
                        ret += dig * ten;
                        ten *= 10;
                }
        }
        return ret;
}
                 /*CHECK FOR LCM*/
 static int lcm(int x, int y)
    {
        int a;
        a = (x > y) ? x : y; // a is greater number
        while(true)
        {
            if(a % x == 0 && a % y == 0)
                return a;
            ++a;
        }	
    }
}