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; }