Time Limit: 1000MS

 

Memory Limit: 262144KB

 

64bit IO Format: %I64d & %I64u

CodeForces - 628C


Bear and String Distance



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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：


#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);
		else
			puts("-1");
	}
	return 0;
}