You are given a string s consisting only of first 20 lowercase Latin letters ('a', 'b', ..., 't').

Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".

You can perform the following operation no more than once: choose some substring s[l; r] and reverse it (i.e. the string s_l s_{l + 1} ... s_r becomes s_r s_{r - 1} ... s_l).

Your goal is to maximize the length of the maximum substring of s consisting of distinct (i.e. unique) characters.

The string consists of distinct characters if no character in this string appears more than once. For example, strings "abcde", "arctg" and "minecraft" consist of distinct characters but strings "codeforces", "abacaba" do not consist of distinct characters.

Input

The only line of the input contains one string s consisting of no more than 10^6 characters 'a', 'b', ..., 't' (first 20 lowercase Latin letters).

Output

Print one integer — the maximum possible length of the maximum substring of s consisting of distinct characters after reversing no more than one its substring.

Examples

Input


abacaba


Output


3


Input


abcdecdf


Output


6


Input


aabbcc


Output


3


Input


abcdeefc


Output


6
import java.util.*;
import java.io.*;
import java.text.*;
public class Main{
    //SOLUTION BEGIN
    //Into the Hardware Mode
    void pre() throws Exception{}
    void solve(int TC)throws Exception{
        String s = n();
        int B = 20, mx = 1<<B;
        int[] f = new int[mx];
        boolean[] g = new boolean[mx];
        for(int i = 0; i< s.length(); i++){
            int mask = 0;
            for(int j = 0; j< 20; j++){
                if(i+j >= s.length())break;
                if(((mask>>(s.charAt(i+j)-'a'))&1)==1)break;
                mask |= 1<<(s.charAt(i+j)-'a');
                f[mask] = bit(mask);
                g[mask] = true;
            }
        }
        for(int b = 0; b < B; b++)
            for(int i = 0; i< mx; i++)
                if(((i>>b)&1)==0)
                    f[i|(1<<b)] = Math.max(f[i|(1<<b)], f[i]);
        int ans = 0;
        for(int i = 0; i< mx; i++){
            if(!g[i])continue;
            ans = Math.max(ans, bit(i)+f[(mx-1)^i]);
        }
        pn(ans);
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    void exit(boolean b){if(!b)System.exit(0);}
    long IINF = (long)1e18, mod = (long)1e9+7;
    final int INF = (int)1e9, MX = (int)2e6+5;
    DecimalFormat df = new DecimalFormat("0.000");
    double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-7;
    static boolean multipleTC = false, memory = false, fileIO = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        if(fileIO){
            in = new FastReader("C:/users/user/desktop/inp.in");
            out = new PrintWriter("C:/users/user/desktop/out.out");
        }else {
            in = new FastReader();
            out = new PrintWriter(System.out);
        }
        //Solution Credits: Taranpreet Singh
        int T = (multipleTC)?ni():1;
        pre();
        for(int t = 1; t<= T; t++)solve(t);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    
    int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}
 
    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
 
        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }
 
        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }
 
        String nextLine() throws Exception{
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }  
            return str;
        }
    }   
}