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:
import java.io.*;
import java.util.Arrays;
import java.util.List;
import java.util.Locale;
import java.util.Map;
import java.util.StringTokenizer;
import java.util.function.Function;
import java.util.function.Supplier;
import java.util.stream.Collector;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

import static java.lang.Math.min;

public class Solution {

    public static final int N = 20;

    boolean[] b = new boolean[1 << N];
    int[] subTreeMax = new int[1 << N];
    boolean[] visited = new boolean[1 << N];

    void solve() throws Throwable {
        char[] a = readString().toCharArray();
        for (int start = 0; start < a.length; start++) {
            int mask = 0;
            for (int i = 0; i < 20; i++) {
                int curInd = start + i;
                if (curInd >= a.length) {
                    break;
                }
                int ch = a[curInd] - 'a';
                if ((mask | (1 << ch)) == mask) {
                    break;
                }
                mask |= (1 << ch);
                b[mask] = true;
            }
        }
        rec((1 << 20) - 1);
        int ans = 0;
        for (int mask = 0; mask < (1 << 20); mask++) {
            ans = Math.max(ans, Integer.bitCount(subTreeMax[mask]) + Integer.bitCount(subTreeMax[((1 << 20) - 1) ^ mask]));
        }
        System.out.println(ans);
    }

    private void rec(int mask) {
        if (visited[mask]) {
            return;
        }
        visited[mask] = true;
        if (b[mask]) {
            subTreeMax[mask] = mask;
        }
        for (int i = 0; i < N; i++) {
            if ((mask & (1 << i)) != 0) {
                int to = mask & ~(1 << i);
                rec(to);
                if (Integer.bitCount(subTreeMax[to]) > Integer.bitCount(subTreeMax[mask])) {
                    subTreeMax[mask] = subTreeMax[to];
                }
            }
        }
    }

    //-------------------------------------------------

    final boolean ONLINE_JUDGE = !new File("input.txt").exists();

    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok;

    public void run() {
        Runnable run = () -> {
            try {
                long startTime = System.currentTimeMillis();
                Locale.setDefault(Locale.US);
                if (ONLINE_JUDGE) {
                    in = new BufferedReader(new InputStreamReader(System.in));
                    out = new PrintWriter(System.out);
                } else {
                    in = new BufferedReader(new FileReader("input.txt"));
                    out = new PrintWriter("output.txt");
                }
                tok = new StringTokenizer("");
                solve();
                in.close();
                out.close();
                long endTime = System.currentTimeMillis();
                long totalMemory = Runtime.getRuntime().totalMemory();
                long freeMemory = Runtime.getRuntime().freeMemory();
                System.err.println();
                System.err.println("Time = " + (endTime - startTime) + " ms");
                //System.err.println("Memory = " + ((totalMemory - freeMemory) / 1024) + " KB");
            } catch (Throwable e) {
                e.printStackTrace(System.err);
                System.exit(-1);
            }
        };
        new Thread(null, run, "run", 256 * 1024 * 1024).start();
        min(0, 0);
    }

    String readString() {
        while (!tok.hasMoreTokens()) {
            String line;
            try {
                line = in.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            if (line == null) return null;
            tok = new StringTokenizer(line);
        }
        return tok.nextToken();
    }

    int readInt() {
        return Integer.parseInt(readString());
    }

    long readLong() throws IOException {
        return Long.parseLong(readString());
    }

    double readDouble() throws IOException {
        return Double.parseDouble(readString());
    }

    void debug(Object... o) {
        if (!ONLINE_JUDGE) {
            System.err.println(Arrays.deepToString(o));
        }
    }

    /*long binpow(long a, long n) {
        long r = 1;
        while (n > 0) {
            if ((n & 1) > 0) {
                r *= a;
            }
            a *= a;
            n /= 2;
        }
        return r;
    }/**/

    long binpow(long a, long n, long mod) {
        long r = 1;
        while (n > 0) {
            if ((n & 1) > 0) {
                r = (r * a) % mod;
            }
            a = (a * a) % mod;
            n /= 2;
        }
        return r;
    }/**/

    static long gcd(long x, long y) {
        return y == 0 ? x : gcd(y, x % y);
    }

    private long[] readLongArray(int n) throws IOException {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = readLong();
        }
        return a;
    }

    private int[] readIntArray(int n) throws IOException {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = readInt();
        }
        return a;
    }

    public static void main(String[] args) {
        new Solution().run();
    }
}