Codeforces CF#628 Education 8 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：
    给出一个长度为n的字符串，定义两个字符串的距离就是对应位的字母在字母表上的距离之和，也就是ae跟ea距离为4+4=8。
    给出k，要求构造一个等长的字母串，使得两串距离恰好为k。

　　

题解：
    很容易想到每一位都尽量拉长距离，尽快达到k，然后全部一样即可。

 

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 100010;
int n, k;
char str[N];
char ans[N];

int main() {
    scanf("%d%d", &n, &k);
    scanf("%s", str);
    memset(ans, 0, sizeof(ans));
    for(int i = 0; i < n; ++i) {
        int dis = min(k, max(str[i] - 'a', 'z' - str[i]));
        if(str[i] - 'a' >= dis) ans[i] = str[i] - dis;
        else ans[i] = str[i] + dis;
        k -= dis;
    }
    if(k) puts("-1");
    else printf("%s\n", ans);
    return 0;
}