Problem:
Score : 300 points

Problem Statement

Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.

A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, atcoder, zscoder and agc are diverse words while gotou and connect aren't diverse words.

Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.

Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.

Constraints

1 \leq |S| \leq 26

S is a diverse word.

Input

Input is given from Standard Input in the following format:

S

Output

Print the next word that appears after S in the dictionary, or -1 if it doesn't exist.

Sample Input 1

atcoder

Sample Output 1

atcoderb

atcoderb is the lexicographically smallest diverse word that is lexicographically larger than atcoder. Note that atcoderb is lexicographically smaller than b.

Sample Input 2

abc

Sample Output 2

abcd

Sample Input 3

zyxwvutsrqponmlkjihgfedcba

Sample Output 3

-1

This is the lexicographically largest diverse word, so the answer is -1.

Sample Input 4

abcdefghijklmnopqrstuvwzyx

Sample Output 4

abcdefghijklmnopqrstuvx

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author ankur
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskA {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            String s = in.readString();

            if (s.length() < 26) {
                for (char i = 'a'; i <= 'z'; i++) {
                    boolean x = false;
                    for (int j = 0; j < s.length(); j++) {
                        if (s.charAt(j) == i) {
                            x = true;
                        }
                    }
                    if (!x) {
                        out.print(s + i);
                        return;
                    }
                }
            }
            for (int i = 25; i >= 1; i--) {
                for (int j = i - 1; j >= 0; j--) {
                    //System.out.println(i+" "+j);
                    if (s.charAt(i) > s.charAt(j)) {
                        char ar[] = in.constring(s);
                        ar[i] = s.charAt(j);
                        ar[j] = s.charAt(i);
                        for (int k = 0; k <= j; k++) {
                            out.print(ar[k]);
                        }
                        return;
                    }
                }
            }
            out.print(-1);
        }

    }

    static class InputReader {
        private final InputStream stream;
        private final byte[] buf = new byte[8192];
        private int curChar;
        private int snumChars;

        public InputReader(InputStream st) {
            this.stream = st;
        }

        public int read() {
            //*-*------clare------
//remeber while comparing 2 non primitive data type not to use ==
//remember Arrays.sort for primitive data has worst time case complexity of 0(n^2) bcoz it uses quick sort
//again silly mistakes ,yr kb tk krta rhega ye mistakes
//try to write simple codes ,break it into simple things
//knowledge>rating
        /*
        public class Main
        implements Runnable{
    public static void main(String[] args) {
        new Thread(null,new Main(),"Main",1<<26).start();

    }
    public void run() {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();//chenge the name of task
        solver.solve(1, in, out);
        out.close();
    }
         */

            if (snumChars == -1)
                throw new InputMismatchException();
            if (curChar >= snumChars) {
                curChar = 0;
                try {
                    snumChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (snumChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        public char[] constring(String str) {
            char[] ar = new char[str.length()];
            for (int i = 0; i < str.length(); i++) {
                ar[i] = str.charAt(i);
            }
            return ar;
        }

        public String readString() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }

        public boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

    }
}