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:
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;
/**
 *
 * Yerasyl Abenov
 *
 */
public class Main {
    public static void main(String[] args) throws IOException {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        //try (PrintWriter out = new PrintWriter(new File("figures.out"))) {
        try (PrintWriter out = new PrintWriter(outputStream)) {
            TaskB solver = new TaskB();
            solver.solve(1, in, out);
        }
    }
}
class TaskB {
    public void solve(int testNumber, InputReader in, PrintWriter out) throws FileNotFoundException, IOException {
        String s = in.next().trim();
        int a[] = new int[26];
        if (s.length() < 26) {
            out.println(-1);
            return;
        }
        int not = 0;
        int cnt = 0;
        int pos = s.length() - 26;
        for (int i = 0; i < 26; ++i) {
            if (s.charAt(i) != '?') {
                int k = s.charAt(i) - 'A';
                if (a[k] == 0)
                    cnt++;
                a[k]++;
            }
            else not++;
        }

        for (int i = 26; i < s.length(); ++i) {
            if (not + cnt == 26) {
                pos = i - 26;
                break;
            }
            if (s.charAt(i - 26) == '?') {
                not--;
            }
            else {
                int k = s.charAt(i - 26) - 'A';
                if (a[k] - 1 == 0) cnt--;
                a[k]--;
            }
            if (s.charAt(i) != '?') {
                int k = s.charAt(i) - 'A';
                if (a[k] == 0) {
                    cnt++;
                }
                a[k]++;
            }
            else not++;
        }
        if (cnt + not == 26) {
            char need[] = new char[not];
            int tmp = 0;
            for (int i = 0; i < 26; ++i) {
                if (a[i] == 0) {
                    need[tmp++] = (char)('A' + i);
                }
            }
            for (int i = 0; i < s.length(); ++i) {
                if (s.charAt(i) == '?') {
                    if (i >= pos && i < pos + 26) out.print(need[--tmp]);
                    else out.print('A');
                }
                else out.print(s.charAt(i));
            }
        }
        else out.println(-1);
    }
}
class InputReader {
    private final BufferedReader reader;
    private StringTokenizer tokenizer;
    public InputReader(InputStream stream) throws FileNotFoundException {
        //reader = new BufferedReader(new FileReader("figures.in"));
        reader = new BufferedReader(new InputStreamReader(stream));
        tokenizer = null;
    }


    public String nextLine() {
        try {
            return reader.readLine();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(nextLine());
        }
        return tokenizer.nextToken();
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong() {
        return Long.valueOf(next());
    }
    public BigInteger nextBigInteger() {
        return new BigInteger(next());
    }
}