Codeforces Round #372 (Div. 2) B

Posted 2020-08-09 樱花落舞

Description

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 asABCDEFGHIJKLMNOPQRSTUVWXYZ 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

题意：给出字符串，如果字符串中有长度为26的子串，且子串中的‘？’经过替换可以刚刚好包含26个字母，就输出整个替换后的字符串

解法：首先小于26的直接-1，然后我们依次截取26长度的字符串进行测试，如果可以替换成功，则另外的‘？’一并取A（可以任意，反正有一个子串已经满足要求了），然后输出整个字符串

另外替换可以用map标记，缺失哪个字母就加哪个，然后标记出现过

#include <bits/stdc++.h>
using namespace std;
string s;
map<int,int>q;
int main()
{
    cin>>s;
    if(s.length()<26)
    {
        cout<<"-1"<<endl;
        return 0;
    }
    else
    {
        string sss="";
        int flag=0;
        string ss;
        int i;
        for(i=0;i<s.length()-25;i++)
        {
            ss=s.substr(i,26);
            flag=0;
         //   cout<<ss<<endl;
            for(int z=0;z<26;z++)
            {
                if(q[ss[z]-‘A‘]&&ss[z]!=‘?‘)
                {
                    flag=1;
                }
                else if(ss[z]==‘?‘)
                {
                    q[27]=1;
                }
                else if(ss[z]<=‘Z‘&&ss[z]>=‘A‘)
                {
                    q[ss[z]-‘A‘]=1;
                }
            }
            //cout<<flag<<"A"<<endl;
            if(flag==0)
            {
                for(int g=0;g<26;g++)
                {
                    if(ss[g]==‘?‘)
                    {
                        for(int h=0;h<26;h++)
                        {
                            if(!q[h])
                            {
                                sss+=(‘A‘+h);
                                //printf("%c",‘A‘+h);
                                q[h]=1;
                                break;
                            }
                        }
                    }
                    else if(ss[g]!=‘?‘)
                    {
                        sss+=ss[g];
                       // cout<<ss[i]<<"^"<<endl;;
                    }
                }
                break;
            }
            else
            {
                q.clear();
            }
           // cout<<flag<<endl;
        }
        if(flag)
        {
            cout<<"-1"<<endl;
        }
        else
        {
            for(int l=0;l<i;l++)
            {
                if(s[l]==‘?‘)
                    cout<<"A";
                else
                    cout<<s[l];
            }
            cout<<sss;
            for(int l=i+26;l<s.length();l++)
            {
                if(s[l]==‘?‘)
                    cout<<"A";
                else
                    cout<<s[l];
            }
        }
    }
    return 0;
}