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.util.*;
import java.io.*;
public class Main {
	public static long mod= (long) (1e9 +7);
	
	public static void main(String args[])
	{
			InputReader s= new InputReader(System.in);
			OutputStream outputStream= System.out;
			PrintWriter out= new PrintWriter(outputStream);
			
			String str= s.nextLine();
			if(str.length()<26)
				out.println("-1");
			else
			{
				int index=0;
				StringBuilder ans=new StringBuilder();
				int flag=0;
				int flag2=0;
				for(int i=0;i<str.length()-25 && flag==0;i++)
				{
					int count[]= new int[100];
					
					for(int j=i;j<i+26;j++)
					{
						if(str.charAt(j)!='?')
							count[(int)str.charAt(j)]++;
					}
					
					for(int j=65;j<=90;j++)
					{
						if(count[j]>1) 
						{
							flag2=1;
							break;
						}
					}
					
					if(flag2==1)
					{
						flag2=0;
						continue;
					}
					
					ArrayList<Character> arr= new ArrayList<>();
					
					for(int k=65;k<=90;k++)
					{
						if(count[k]==0)
						{
							arr.add((char)k);
						}
					}
					
					int temp=0;
					for(int j=i;j<i+26;j++)
					{
						if(str.charAt(j)!='?')
							ans.append(str.charAt(j));
						else
						{
							ans.append(arr.get(temp));
							temp++;
						}
						index=i;
						flag=1;
					}
				}
				//System.out.println(ans);
				if(ans.length()==0)
				{
					out.println("-1");
				}
				else
				{
					StringBuilder ans1=new StringBuilder();
					for(int i=0;i<index;i++)
					{
						if(str.charAt(i)=='?') 
							ans1.append('A');
						else 
							ans1.append(str.charAt(i));
					}
					ans1.append(ans);
					for(int i=index+26;i<str.length();i++)
					{
						if(str.charAt(i)=='?') 
							ans1.append('A');
						else 
							ans1.append(str.charAt(i));
					}
					out.print(ans1);
				}
			}
			out.close();
	}
	static class InputReader {

	    public BufferedReader reader;
	    public StringTokenizer tokenizer;
 
	    public InputReader(InputStream inputstream) {
	      reader = new BufferedReader(new InputStreamReader(inputstream));
	      tokenizer = null;
	    }
	    
	    public String nextLine(){
	    	String fullLine=null;
	    	while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	              fullLine=reader.readLine();
	            } catch (IOException e) {
	              throw new RuntimeException(e);
	            }
	            return fullLine;
	          }
	          return fullLine;
	    }
		public String next() {
	      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	        try {
	          tokenizer = new StringTokenizer(reader.readLine());
	        } catch (IOException e) {
	          throw new RuntimeException(e);
	        }
	      }
	      return tokenizer.nextToken();
	    }
		public long nextLong() {
		      return Long.parseLong(next());
		    }
	    public int nextInt() {
	      return Integer.parseInt(next());
	    }
	  }
	static class Pair implements Comparable<Pair>
	{
		long f,s;
		Pair(long ii, long cc)
		{
			f=ii;
			s=cc;
		}
		
		public int compareTo(Pair o) 
		{
			return Long.compare(this.f, o.f);
		}
		
	}
	public static int gcd(int number1, int number2) {  
		if(number2 == 0){
			return number1;
		} 
		return gcd(number2, number1%number2); 
	}
	public static int combinations(int n,int r){
		if(n==r) return 1;
		if(r==1) return n;
		if(r==0) return 1;
		return combinations(n-1,r)+ combinations(n-1,r-1);
	}
	public static long binomialCoeff(int n, int k)
	{
	    long C[][]= new long[n+1][k+1];
	    int i, j;
	 
	    // Caculate value of Binomial Coefficient in bottom up manner
	    for (i = 0; i <= n; i++)
	    {
	        for (j = 0; j <= Math.min(i, k); j++)
	        {
	            // Base Cases
	            if (j == 0 || j == i)
	                C[i][j] = 1;
	 
	            // Calculate value using previosly stored values
	            else
	                C[i][j] = C[i-1][j-1] + C[i-1][j];
	        }
	    }
	 
	    return C[n][k];
	}
	public static int expo(int a, int b){
	    if (b==1)
	        return a;
	    if (b==2)
	        return a*a;

	    if (b%2==0){
	            return expo(expo(a,b/2),2);
	    }
	    else{
	        return a*expo(expo(a,(b-1)/2),2);
	    }
	}
	
	public static void sieve(int N){
		int arr[]= new int[N+1];
		for(int i=2;i<Math.sqrt(N);i++){
			if(arr[i]==0){
				for(int j= i*i;j<= N;j= j+i){
					arr[j]=1;
				}
			}
		}
		// All the i for which arr[i]==0 are prime numbers.
	}
}