CodeForces 628 C C - Bear and String Distance

C - Bear and String Distance

Time Limit:1000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

SubmitStatusPracticeCodeForces 628C

Description

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,, and.

Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example,, and.

Limak gives you a nice string s and an integer k. He challenges you to find any nice strings' that . Find anys' 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 .

Sample Input

Input

4 26
bear

Output

roar

Input

2 7
af

Output

db

Input

3 1000
hey

Output

-1


代码如下：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
char a[100005];//原字符串
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        scanf("%s",a);
        int sum=0;
        for(int i=0;i<n;i++)
            sum+=max(a[i]-'a','z'-a[i]);
        if(sum<k)
            printf("-1\n");
        else
        {
            int p,q;
            for(int i=0;i<n;i++)
            {
                p=a[i]-'a',q='z'-a[i];
                if(!k)
                    cout << a[i];
                else if(max(p,q)>k)
                {
                    printf("%c",a[i]+k > 'z'?a[i]-k:a[i]+k);
                    k=0;
                }
                else
                {
                    printf("%c",p>q?'a':'z');
                    k-=max(p,q);
                }
            }
        }
    }
}