Dean and Schedule （URAL 2026）

Problem

A new academic year approaches, and the dean must make a schedule of classes for first-year students. There must be n classes in the schedule. The dean should take into account the following interesting observation made in recent years: students skip all classes with even numbers and attend all classes with odd numbers (the classes are numbered from 1). Of course the dean wants students to attend as many important classes as possible, so he tries to assign subjects that are more important to places with odd numbers and subjects that are less important to places with even numbers. The method of estimating the quality of the schedule at the Department of Mathematics and Mechanics must be as formal as possible.

The first-year schedule may contain any of 26 subjects taught at the department. We denote them by English letters from a to  z. The importance of a subject corresponds to its position in the English alphabet. Thus, subject a has importance 1, and subject  z has importance 26. The quality of a schedule is the sum of importances of subjects in it, where subjects on odd places are counted with a plus sign, and subjects on even places are counted with a minus sign.

Unfortunately, a shedule has some restrictions due to administrative reasons. First, the schedule must contain at least k different subjects, so the dean cannot occupy all odd places with mathematical analysis and all even places with philosophy. Second, certain subjects must be assigned to certain places. Help the dean to make a schedule of maximum quality under these restrictions.

Input

The first line contains a string of length n (1 ≤ n ≤ 10 5) consisting of lowercase English letters and question marks. The string specifies the subjects that are already in the schedule. The letters denote these subjects, and the question marks stand for vacant places. In the second line you are given an integer k (1 ≤ k≤ 26), which is the minimum number of different subjects in the schedule.

Output

If it is impossible to replace all question marks by lowercase English letters so that the string would contain at least k different letters, output “-1” (without quotation marks). Otherwise, output any of the resulting strings that maximizes the quality of the schedule given by the string.

Example

input	output
?? 1	za
?? 3	-1
aza 1	aza
aza 3	-1

Notes

In the first sample the dean can make any schedule with two subjects (even identical), but the quality of the schedule “za” is 26 − 1 = 25, and this is the maximum possible value of the quality.

In the second sample it is impossible to make a schedule consisting of two classes with three different subjects.

In the third sample the dean has only one variant. Though the schedule is bad (1 − 26 + 1 = −24), nothing better can be proposed.

In the fourth sample the only possible variant doesn’t contain three different subjects.

题解：（抽象理解一下qwq）要安排课程，课程的种类是a~z，每个课程都有不同的权值，a~z对应1~26，如果在奇数位上就是加上这个权值，如果是偶数位，就减去这个权值，现在给你一串字符串，里面有字母和问号，问号是由你来安排，字母是已经安排好的不需要再改变，再给你一个数k，问能够安排大于k种类的课程，且要求权值最大。

只需要找出来添加的种类，添加的字符只添加一个，并且是偶数位就从a开始找，是奇数就从z开始找，其他位置是奇数位就输出z，是偶数就输出a。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
using namespace std;
typedef long long ll;
char a[100005];       // a是读入的字符串
int num[505] = {0};   // 用来统计那个字符出现过
int main()
{
    int n, i, j, k;
    scanf("%s %d", &a, &k);
    int len = strlen(a);
    int sum1 = 0, sum2 = 0, n1 = 0, n2 = 0;   //sum1是问号的个数 sum2是出现字符种类
    for(i = 0; i < len; i++)                 // n1记录出现的？在奇数位的个数，n2记录在偶数位
    {
        if(a[i] >= 'a' && a[i] <= 'z')num[a[i]]++;
        else sum1++;
        if(a[i] == '?')
        {
            if(i % 2 == 0)n1++;
            else n2++;
        }
    }
    for(i = 'a'; i <= 'z'; i++)  // 记录出现的字符个数
    {
        if(num[i])sum2++;
    }
    if(sum2 + sum1 < k)  // 如果不足够k 则不满足情况
    {
        printf("-1\n");
        return 0;
    }
    int sum = max(k - sum2, 0);   // 记录还需要添加的种类
    for(i = 0; sum != 0; i++)     // 通过去掉需要添加的，判断可以任意安排的个数
    {
        if(num['a' + i] == 0 && n2 > 0)   // 从两边开始寻找，那么差值一定最小，即权值一定最大
        {
            n2--;
            sum--;
            if(sum == 0)break;
        }
        if(num['z' - i] == 0 && n1 > 0)
        {
            n1--;
            sum--;
            if(sum == 0)break;
        }
    }
    for(i = 0; i < len; i++)  // 处理、输出
    {
        if(a[i] != '?')printf("%c", a[i]);  // 如果已经安排好的，直接输出
        else
        {
            if(i % 2 == 0)   // 如果是奇数位
            {
                if(n1 > 0)   //如果还有可以自由安排的奇数位数的地方，输出z,权值高
                {
                    printf("z");
                    n1--;
                }
                else        // 输出任意之后，输出需要的种类
                {
                    for(j = 'z'; j >= 'a'; j--)
                    {
                        if(num[j] == 0)break;   // 因为奇数位相加，所以从最大的开始遍历，如果原来不存在则输出
                    }
                    num[j]++;   // 输出后标记
                    printf("%c", j);
                }
            }
            else
            {
                if(n2 > 0)   // 同理找偶数位
                {
                    printf("a");
                    n2--;
                }
                else
                {
                    for(j = 'a'; j <= 'z'; j++)
                    {
                        if(num[j] == 0)break;
                    }
                    num[j]++;
                    printf("%c", j);
                }
            }
        }
    }
    printf("\n");
    return 0;
}