Problem:
Score : 600 points

Problem Statement

A subsequence of a string S is a string that can be obtained by deleting zero or more characters from S without changing the order of the remaining characters.
For example, arc, artistic and (an empty string) are all subsequences of artistic; abc and ci are not.

You are given a string A consisting of lowercase English letters.
Find the shortest string among the strings consisting of lowercase English letters that are not subsequences of A.
If there are more than one such string, find the lexicographically smallest one among them.

Constraints

1 \leq |A| \leq 2 \times 10^5

A consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

A

Output

Print the lexicographically smallest string among the shortest strings consisting of lowercase English letters that are not subsequences of A.

Sample Input 1

atcoderregularcontest

Sample Output 1

b

The string atcoderregularcontest contains a as a subsequence, but not b.

Sample Input 2

abcdefghijklmnopqrstuvwxyz

Sample Output 2

aa

Sample Input 3

frqnvhydscshfcgdemurlfrutcpzhopfotpifgepnqjxupnskapziurswqazdwnwbgdhyktfyhqqxpoidfhjdakoxraiedxskywuepzfniuyskxiyjpjlxuqnfgmnjcvtlpnclfkpervxmdbvrbrdn

Sample Output 3

aca

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {

    static String A;

    public static void main(String[] args) {
        FastScanner sc = new FastScanner(System.in);
        A = sc.next();

        System.out.println( solve() );
    }

    static String solve() {
        long prev = 0;
        int cur = 1;
        // distsは広義単調減少になる
        int[] dists = new int[A.length()];
        for (int i = A.length()-1; i >= 0; i--) {
            char c = A.charAt(i);
            dists[i] = cur;
            int cn = c - 'a';

            prev |= 1 << cn;

            if( prev == (1 << 26) - 1 ) {
                prev = 0;
                cur++;
            }
        }

        StringBuilder ans = new StringBuilder();
        int chars = 0;
        int[] charDist = new int[26];
        int[] charDistIndex = new int[26];
        Arrays.fill(charDist, -1);
        for (int i = 0; i < A.length(); i++) {
            char c = A.charAt(i);
            int dist = dists[i];

            if( charDist[c - 'a'] == -1 ) {
                charDist[c - 'a'] = dist;
                charDistIndex[c - 'a'] = i;
                chars++;
            }

            // 全部出そろった時点で判定する
            if( chars == 26 ) {
                for (int j = 0; j < 26; j++) {
                    if( charDist[j] == dist ) { // 広義単調減少なので
                        char a = (char)('a' + j);
                        ans.append(a);
                        i = charDistIndex[j]; // そこまで巻き戻す
                        break;
                    }
                }

                chars = 0;
                Arrays.fill(charDist, -1);
                Arrays.fill(charDistIndex, -1); // こっちはほっといてもよいけど一応
            }
        }

        // 最後に出てないものを利用する
        for (int i = 0; i < 26; i++) {
            if( charDist[i] == -1 ) {
                ans.append((char)(i + 'a'));
                break;
            }
        }

        return ans.toString();
    }

    @SuppressWarnings("unused")
    static class FastScanner {
        private BufferedReader reader;
        private StringTokenizer tokenizer;

        FastScanner(InputStream in) {
            reader = new BufferedReader(new InputStreamReader(in));
            tokenizer = null;
        }

        String next() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        String nextLine() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    return reader.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken("\n");
        }

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

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }
    }
}