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.

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.

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 orABCEDFGHIJKLMNOPQRZTUVWXYS.

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 26that contains all letters of the English alphabet fits as an answer.

题目巴拉巴拉很多 有用的信息不多

题意：给你一个字符串 有没有一个有着连续的长度为26的字串，它包含所有的26个字母

这题就是纯暴力，从第 i 个字符开始搜索看看i+26 符不符合条件

 #include<cstdio>

 #include<cstring>

 #include<cmath>

 #include<algorithm>

 #include<queue>

 using namespace std;

 char a[];

 int b[];

 int main() {

     while(scanf("%s",a)!=EOF) {

         int n=strlen(a),flag1,flag2=;

         if (n<) {

             printf("-1\n");

             continue;

         }

         for (int i= ; i<n- ; i++ ) {

             flag1=;

             memset(b,,sizeof(b));

             for (int j=i ; j<i+ ; j++) {

                 if (a[j]>='A' && a[j]<='Z') {

                     b[a[j]-'A']++;

                 }

                 if  (b[a[j]-'A']>=) {

                     flag1=;

                     break;

                 }

             }

             if (!flag1) continue;

             flag2=;

             for (int j=i ; j<i+ ; j++) {

                 if (a[j]=='?') {

                     for (int k= ; k< ; k++) {

                         if (b[k]==) {

                             a[j]=k+'A';

                             b[k]=;

                             break;

                         }

                     }

                 }

             }

             if (flag2) break;

         }

         if (flag2) {

             for (int i= ;i<n ;i++){

                 if (a[i]=='?') printf("A");

                 else printf("%c",a[i]);

             }

             printf("\n");

         }else printf("-1\n");

     }

     return ;

 }