Problem: 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

Solution (in JAVA):

Solution:
// Don't place your source in a package

import java.math.BigInteger;
import java.util.*;
import java.lang.*;
import java.io.*;



// Please name your class Main
public class Main {
    static FastScanner fs=new FastScanner();
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        int Int() {
            return Integer.parseInt(next());
        }

        long Long() {
            return Long.parseLong(next());
        }

        String Str(){
            return next();
        }
    }


    public static void main (String[] args) throws java.lang.Exception {
        PrintWriter out = new PrintWriter(System.out);


        int T=1;
        for(int t=0;t<T;t++){
            String s=Str();

            Solution sol=new Solution(out);
            sol.solution(s);
        }
        out.flush();

    }


    public static int Int(){
        return fs.Int();
    }
    public static long Long(){
        return fs.Long();
    }
    public static String Str(){
        return fs.Str();
    }

}






class Solution{
    PrintWriter out;
    public Solution(PrintWriter out){
        this.out=out;
    }


    int dp[];
    int bits[];
    public void solution(String s){


        int n=s.length();
        bits=new int[(1<<21)+10];
        dp=new int[bits.length];
        Arrays.fill(dp,-1);
        for(int i=0;i<n;i++){
            int bit=0;
            for(int j=i;j<n;j++){
                int d=s.charAt(j)-'a';
                if((bit&(1<<d))!=0)break;
                bit=bit|(1<<d);
                bits[bit]=Math.max(j-i+1,bits[bit]);
            }
        }

        int state=(1<<21)-1;
        dfs(state);

        int res=0;

        for(int i=0;i<state;i++){
            if(state-i<0)break;
            res=Math.max(res,bits[i]+dp[state-i]);
        }
        out.println(res);
    }

    public int dfs(int state){
        if(state==0)return 0;

        if(bits[state]!=0){
            dp[state]=Math.max(bits[state],dp[state]);
            return bits[state];
        }
        if(dp[state]!=-1)return dp[state];

        int res=0;
        for(int i=0;i<21;i++){
            if((state&(1<<i))!=0){
                res=Math.max(res,dfs(state^(1<<i)));
            }
        }
        dp[state]=res;
        return res;
    }







}



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