codeforces 628C C. Bear and String Distance-优快云博客

本文介绍了一个关于字符串距离的问题，即找到一个字符串使得它与给定字符串之间的字母距离总和等于指定值。文章提供了AC代码，并解释了如何通过调整每个字符来达到目标距离。

Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.

The distance between two letters is defined as the difference between their positions in the alphabet. For example, $\text{[math]}$ , and $\text{[math]}$ .

Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, $\text{[math]}$ , and $\text{[math]}$ .

Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that $\text{[math]}$ . Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to usegets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead ofScanner/System.out in Java.

Input

The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).

The second line contains a string s of length n, consisting of lowercase English letters.

Output

If there is no string satisfying the given conditions then print "-1" (without the quotes).

Otherwise, print any nice string s' that $\text{[math]}$ .

Examples

input

4 26
bear

output

roar

input

2 7
af

output

db

input

3 1000
hey

output

-1

题意：

AC代码：

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+4;
char s[N];
int n,k;
int main()
{
    int num=0;
    scanf("%d%d",&n,&k);
    scanf("%s",s);
    for(int i=0;i<n;i++)
    {
        num+=max(s[i]-'a','z'-s[i]);
    }
    if(k>num)cout<<"-1"<<"\n";
    else
    {
        int sum;
        for(int i=0;i<n;i++)
        {
            sum=max(s[i]-'a','z'-s[i]);
            //cout<<sum<<endl;
            if(sum<=k){
                    k-=sum;
            if(s[i]-'a'>'z'-s[i])s[i]='a';
            else s[i]='z';}
            else if(sum>k&&k!=0)
            {
                if(s[i]-'a'>=k)
                {
                    s[i]=s[i]-k;
                }
                else
                {
                    s[