Codeforces Round #372 (Div. 2) B. Complete the Word（模拟）

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 length26 where each letter of English alphabet appears exactly once. In particular, if the string has length strictly less than26, 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 length26 (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 orABCEDFGHIJKLMNOPQRZTUVWXYS.

In the second sample case, there are no missing letters. In addition, the given string does not have a substring of length26 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：给出一个字符串，问是否含有一个子串，这个子串只有26个字符，并且包含A-Z的所有字母，字符串中的 ‘？’ 可以用任何大写字母来代替，不包含则输出-1

思路：从第一个字符开始模拟即可，一直到len-26，内循环检测子串这26个字符，出现的字母用book标记一下，并且记录问号的个数，如果问号的个数等于这26个字符缺少的字母数，说明存在符合题意的子串，把缺少的字母记录下来，然后替换到？的位置，注意要求输出整个字符串，并且没有问号出现，开始我只输出的符合要求的子串wa了好几次，对于这种模拟题写的时候一定要仔细，要把各种细节都要考虑到

#include<bits/stdc++.h>using namespace std;char s[500005];int main(void){scanf("%s",s);int len=strlen(s);int book[30];int flag=0;for(int i=0;i<=len-26;i++){int count_w=0;//问号的个数memset(book,0,sizeof(book));for(int j=i;j<i+26;j++){if(s[j]=='?'){count_w++; }else{//进行标记 book[(int)(s[j])-(int)('A')]=1;}} int les=0;//缺少字母的个数char less[30];//记录缺少的字母 for(int k=0;k<26;k++){if(book[k]==0){less[les++]=(char)(k+(int)('A'));}}//当缺少的字母个数与问号的个数相等时，说明//可以产生一个序列 if(les==count_w){flag=1;int m=0;for(int j=i;j<i+26;j++){if(s[j]=='?'){//将问号用缺少的字母代替 s[j]=less[m++];}}for(int n=0;n<len;n++){if(s[n]=='?'){s[n]='A';//把问号用任意大写字母代替即可 }} }}if(flag){printf("%s",s);}else{printf("-1\n");}return 0;}