CodeForces - 628C Bear and String Distance （模拟）

CodeForces - 628C

Bear and String Distance

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

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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 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 use gets/scanf/printf instead of getline/cin/cout in C++, prefer to useBufferedReader/PrintWriter instead 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 .

Sample Input

Input

4 26bear

Output

roar

Input

2 7af

Output

db

Input

3 1000hey

Output

-1

//题意：

定义dist(字母1， 字母2) = 两字母ascll之差。
给定一个串a和一个值k，找到另一个串b(长度相等)使得两串各个位置的dist()之和 == k。若得不到k输出-1
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> #include<set> using namespace std; #define mem(x,y) memset(x,y,sizeof(x))const int MAXN = 1e5 + 100; char s[MAXN];int main(){int n, k;while(~scanf("%d%d", &n, &k)){scanf("%s", s);for(int i = 0; i < n; i++){int t = s[i] - 'a';int d = 25 - t;if(t >= d){if(k >= t){s[i] = 'a';k -= t;}else{s[i] = s[i] - k;k -= k;break;}}else{if(k >= d){s[i] = 'z';k -= d;}else{s[i] += k;k -= k;break;}}}if(k == 0)printf("%s\n", s);elseputs("-1");}return 0;}