ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists asubstring (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.

Example

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 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 ;
}

Complete the Word CodeForces - 716B的更多相关文章

  1. Codeforces 716B Complete the Word【模拟】 (Codeforces Round #372 (Div. 2))

    B. Complete the Word time limit per test 2 seconds memory limit per test 256 megabytes input standar ...

  2. Codeforces Round #372 (Div. 2) A .Crazy Computer/B. Complete the Word

    Codeforces Round #372 (Div. 2) 不知不觉自己怎么变的这么水了,几百年前做A.B的水平,现在依旧停留在A.B水平.甚至B题还不会做.难道是带着一种功利性的态度患得患失?总共 ...

  3. codeforces 372 Complete the Word(双指针)

    codeforces 372 Complete the Word(双指针) 题链 题意:给出一个字符串,其中'?'代表这个字符是可变的,要求一个连续的26位长的串,其中每个字母都只出现一次 #incl ...

  4. B. Complete the Word(Codeforces Round #372 (Div. 2)) 尺取大法

    B. Complete the Word time limit per test 2 seconds memory limit per test 256 megabytes input standar ...

  5. CodeForces 716B Complete the Word

    题目链接:http://codeforces.com/problemset/problem/716/B 题目大意: 给出一个字符串,判断其是否存在一个子串(满足:包含26个英文字母且不重复,字串中有‘ ...

  6. Complete the Word

    ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring ...

  7. CSUST 8.5 早训

    ## Problem A A - Meeting of Old Friends CodeForces - 714A 题意: 解题说明:此题其实是求两段区间的交集,注意要去除掉交集中的某个点. 题解: ...

  8. Codeforces水题集合[14/未完待续]

    Codeforces Round #371 (Div. 2) A. Meeting of Old Friends |B. Filya and Homework A. Meeting of Old Fr ...

  9. Codeforces Round #370 - #379 (Div. 2)

    题意: 思路: Codeforces Round #370(Solved: 4 out of 5) A - Memory and Crow 题意:有一个序列,然后对每一个进行ai = bi - bi  ...

随机推荐

  1. 常系数齐次线性递推 & 拉格朗日插值

    常系数齐次线性递推 具体记在笔记本上了,以后可能补照片,这里稍微写一下,主要贴代码. 概述 形式: \[ h_n = a_1 h_{n-1}+a_2h_{n-2}+...+a_kh_{n-k} \] ...

  2. JAVAEE——BOS物流项目03:学习计划、messager、menubutton、登陆拦截器、信息校验和取派员添加功能

    1 学习计划 1.jQuery easyUI messager使用方式 n alert方法 n confirm方法 n show方法 2.jQuery easyUI menubutton使用方式 3. ...

  3. Windows Server 2016-FSMO操作主机角色介绍

    FSMO五个操作主机角色 1.林范围操作主机角色(两种): 架构主机角色:Schema Master 域命名主机角色:Domain Naming Master 2.域范围操作主机角色(三种): 域范围 ...

  4. ES6 学习笔记之二 块作用域与闭包

    "闭包是函数和声明该函数的词法环境的组合." 这是MDN上对闭包的定义. <JavaScript高级程序设计>中则是这样定义的:闭包是指有权访问另一个函数作用域中的变量 ...

  5. CSS3 @keyframes 用法(简单动画实现)

    定义: 通过 @keyframes 规则,能够创建动画. 创建动画的原理是,将一套 CSS 样式逐渐变化为另一套样式. 在动画过程中,可以多次改变这套 CSS 样式. 以百分比来规定改变发生的时间,或 ...

  6. PHPUnit-附录 A. 断言 (assert)

    [http://www.phpunit.cn/manual/5.7/zh_cn/appendixes.assertions.html] 本附录列举可用的各种断言方法. assertArrayHasKe ...

  7. 用最简单的例子实现jQuery图片即时上传

    [http://www.cnblogs.com/Zjmainstay/archive/2012/08/09/jQuery_upload_image.html] 最近看了一些jQuery即时上传的插件, ...

  8. Linux下LNMP启动不了的问题总结(2015.05)

    [1] *****@*****-VirtualBox:~$ sudo /etc/init.d/mysql.server start Starting MySQL * Couldn't find MyS ...

  9. vue框架-学习记录

    前段时间在做vue项目时,遇到挺多问题,想简单总结一下: 1.关于父组件,子组件的通信 网上有很多这方面的讲解,讲解也比较细致,我主要总结了自己在项目中需要的: [1]父组件-子组件 也就是" ...

  10. 一个例子理解break和continue的区别

    结论:break用于终止整个循环,而continue用于终止某一次循环.public class Test { public static void main(String[] args) { for ...