A. Problem Generator

Felven2024-06-13 9:55

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vlad is planning to hold m𝑚 rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of n𝑛 problems, where the i𝑖-th problem has a difficulty level of ai𝑎𝑖. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold m𝑚 rounds.

For example, if m=1𝑚=1, n=10𝑛=10, a=𝑎= 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

Input

The first line contains a single integer t𝑡 (1≤t≤10001≤𝑡≤1000) --- the number of test cases.

The first line of each test case contains two integers n𝑛 and m𝑚 (1≤n≤501≤𝑛≤50, 1≤m≤51≤𝑚≤5) --- the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string a𝑎 of n𝑛 characters from 'A' to 'G' --- the difficulties of the problems in the bank.

Output

For each test case, output a single integer --- the minimum number of problems that need to come up with to hold m𝑚 rounds.

Example

input

Copy

复制代码

10 1

BGECDCBDED

10 2

BGECDCBDED

9 1

BBCDEFFGG

output

Copy

复制代码

2
5
1

解题说明：此题是一道模拟器，因为每一轮都必须要出现A-G这几个字母，所以直接统计当前已出现的字母，然后判断每一轮中是否漏字母即可，最后求和。

cpp 复制代码

#include<stdio.h>

int main()
{
	int test;
	scanf("%d", &test);
	while (test--)
	{
		int a;
		int temp[7] = { 0 };
		int b;
		scanf("%d%d", &a, &b);
		char t[55];
		scanf(" %s", t);
		for (int i = 0; i < a; i++)
		{
			temp[t[i] - 65]++;
		}
		int count = 0;
		for (int i = 0; i < 7; i++)
		{
			if ((temp[i]) < b)
			{
				count += (b - (temp[i]));
			}
		}
		printf("%d\n", count);
	}
	return 0;
}