codeforces 628C Bear and String Distance

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/PrintWriterinstead of Scanner/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

定义两个小写字母之间的距离为这两个字母在字母表中的距离，如dis(a,z)=25,dis(a,c)=2，两个长度相同串的距离为这两个串对应位置字母距离之和。现给出一个长度为n的数字串s和一个距离k，问是否存在一个长度为n的串ss，使得dis(s,ss)=k，如果存在任意输出一解，如果不存在则输出-1 

贪心，先找到该串与其他串的距离上限，如果该上限大于k则输出-1，否则每次贪心的构造ss，即ss的每个字母与s相对应字母的距离尽量大，一直到抵消掉k 

#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 100000 + 10;
int main()
{
    char s[maxn];
    int n, k;
    while (~scanf("%d%d%s", &n, &k, s)){
        int num = 0;
        for (int i = 0; i < n; i++)
            num += max(s[i] - 'a', 'z' - s[i]);
        if (num < k)
            printf("-1\n");
        else{
            for (int i = 0; i < n; i++){
                int d1 = s[i] - 'a', d2 = 'z' - s[i];
                if (k == 0)
                    printf("%c", s[i]);
                else if (k < max(d1, d2)){
                    printf("%c", s[i] + k > 'z' ? s[i] - k : s[i] + k);
                    k = 0;
                }
                else{
                    printf("%c", d1 > d2 ? 'a' : 'z');
                    k -= max(d1, d2);
                }

            }
            printf("\n");
        }
    }
    return 0;
}