Problem: ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 26 where each letter of English alphabet appears exactly once. In particular, if the string has length strictly less than 26, no such substring exists and thus it is not nice.

Now, ZS the Coder tells you a word, where some of its letters are missing as he forgot them. He wants to determine if it is possible to fill in the missing letters so that the resulting word is nice. If it is possible, he needs you to find an example of such a word as well. Can you help him?

Input

The first and only line of the input contains a single string s (1 ≤ |s| ≤ 50 000), the word that ZS the Coder remembers. Each character of the string is the uppercase letter of English alphabet ('A'-'Z') or is a question mark ('?'), where the question marks denotes the letters that ZS the Coder can't remember.

Output

If there is no way to replace all the question marks with uppercase letters such that the resulting word is nice, then print  - 1 in the only line.

Otherwise, print a string which denotes a possible nice word that ZS the Coder learned. This string should match the string from the input, except for the question marks replaced with uppercase English letters.

If there are multiple solutions, you may print any of them.

Examples

Input

ABC??FGHIJK???OPQR?TUVWXY?


Output

ABCDEFGHIJKLMNOPQRZTUVWXYS

Input

WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO


Output

-1

Input

??????????????????????????


Output

MNBVCXZLKJHGFDSAQPWOEIRUYT

Input

AABCDEFGHIJKLMNOPQRSTUVW??M


Output

-1

Note

In the first sample case, ABCDEFGHIJKLMNOPQRZTUVWXYS is a valid answer beacuse it contains a substring of length 26 (the whole string in this case) which contains all the letters of the English alphabet exactly once. Note that there are many possible solutions, such as ABCDEFGHIJKLMNOPQRSTUVWXYZ or ABCEDFGHIJKLMNOPQRZTUVWXYS.

In the second sample case, there are no missing letters. In addition, the given string does not have a substring of length 26 that contains all the letters of the alphabet, so the answer is  - 1.

In the third sample case, any string of length 26 that contains all letters of the English alphabet fits as an answer.

Solution (in JAVA):

Solution:
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

import java.awt.Point;
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.text.*;

/**
 *
 * @author zulkan
 */
public class cf716b {

    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter out = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);

    public static String getString() {
        try {
            return br.readLine();
        } catch (Exception e) {
        }
        return "";
    }

    public static Integer getInt() {
        try {
            return Integer.parseInt(br.readLine());
        } catch (Exception e) {
        }
        return 0;
    }

    public static Integer[] getIntArr() {
        try {
            StringTokenizer temp = new StringTokenizer(br.readLine());
            int n = temp.countTokens();
            Integer temp2[] = new Integer[n];
            for (int i = 0; i < n; i++) {
                temp2[i] = Integer.parseInt(temp.nextToken());
            }
            return temp2;
        } catch (Exception e) {
        }
        return null;
    }

    public static Long[] getLongArr() {
        try {
            StringTokenizer temp = new StringTokenizer(br.readLine());
            int n = temp.countTokens();
            Long temp2[] = new Long[n];
            for (int i = 0; i < n; i++) {
                temp2[i] = Long.parseLong(temp.nextToken());
            }
            return temp2;
        } catch (Exception e) {
        }
        return null;
    }

    public static String[] getStringArr() {
        try {
            StringTokenizer temp = new StringTokenizer(br.readLine());
            int n = temp.countTokens();
            String temp2[] = new String[n];
            for (int i = 0; i < n; i++) {
                temp2[i] = (temp.nextToken());
            }
            return temp2;
        } catch (Exception e) {
        }
        return null;
    }

    public static void print(Object a) {
        out.print(a);
    }

    public static void println(Object a) {
        out.println(a);
    }

    public static void printf(String s, Object... a) {
        out.printf(s, a);
    }

    public static int nextInt() {
        return sc.nextInt();
    }

    public static double nextDouble() {
        return sc.nextDouble();
    }

    public static void main(String[] ar) {
        cf716b c = new cf716b();
        c.solve();
        out.flush();
    }

    public void solve() {
        String ors = getString();
        char in[] = ors.toCharArray();
        int n = in.length;
        HashSet<Integer> sli[] = new HashSet[300];
        for (int i = 0; i < sli.length; i++) {
            sli[i] = new HashSet<>();
        }

        for (int i = 0; i < n; i++) {
            char p = in[i];
            if (i < 25) {
                sli[(int) p].add(i);
            } else {
                sli[(int) p].add(i);

                if (i > 25) {
                    char sblm = in[i - 26];
                    sli[(int) sblm].remove(i - 26);
                }
                boolean all1 = true;
                for (char j = 'A'; j <= 'Z'; j++) {
                    all1 &= sli[(int) j].size() <= 1;
                   
                }
                //println("IS all1 " +all1);
                if (all1) {
                    char ou[] = ors.toCharArray();

                    List<Character> ygBlm = new ArrayList<>();

                    for (char j = 'A'; j <= 'Z'; j++) {
                        if(sli[(int) j].size() == 0) {
                            ygBlm.add(j);
                            //println("BELUM "+j);
                        }
                    }
                    
                    
                    int j = 0;
                   
                    for(int pos : sli['?']) {
                        ou[pos] = ygBlm.get(j++);
                        //println("MASUKKAN "+ou[pos]);
                    }
                    
                    for(int k=0;k<n;k++) {
                        if(ou[k] == '?') {
                            ou[k] = 'A';
                        }
                    }
                    
                    println(new String(ou));

                    return;
                }

            }
        }

        println(-1);

    }

}