codeforces 628C C. Bear and String Distance

C. Bear and String Distance

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 string s' that . 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 .

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[i]=s[i]+k;
                }
                k=0;
            }
            printf("%c",s[i]);
        }

    }
    return 0;
}