CodeForces 721B Passwords

http://codeforces.com/problemset/problem/721/B

B. Passwords

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.

Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.

Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.

Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).

Input

The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.

The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.

The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords.

Output

Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.

Examples

input

5 2
cba
abc
bb1
abC
ABC
abc

output

1 15

input

4 100
11
22
1
2
22

output

3 4

Note

Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length2 and only then the right one, spending 4 seconds at all.

不要被 he,不要被hei........

题意：

键入 n 行密码，最后给定正确的密码。每次键入错 k 次后需要等待 5 秒后才能继续键入，

问键入密码成功的最短和最长时间。

不明白是不是密码的键入是默认字符串有长到短。

思路：

记录每个长度字符串的个数，记住的就是：长度有短到长键入，最少次数和最多次数见代码

CODE:

/*
                              _,add8ba,
                            ,d888888888b,
                           d8888888888888b                        _,ad8ba,_
                          d888888888888888)                     ,d888888888b,
                          I8888888888888888 _________          ,8888888888888b
                __________`Y88888888888888P"""""""""""baaa,__ ,888888888888888,
            ,adP"""""""""""9888888888P""^                 ^""Y8888888888888888I
         ,a8"^           ,d888P"888P^                           ^"Y8888888888P'
       ,a8^            ,d8888'                                     ^Y8888888P'
      a88'           ,d8888P'                                        I88P"^
    ,d88'           d88888P'                                          "b,
   ,d88'           d888888'                                            `b,
  ,d88'           d888888I                                              `b,
  d88I           ,8888888'            ___                                `b,
 ,888'           d8888888          ,d88888b,              ____            `b,
 d888           ,8888888I         d88888888b,           ,d8888b,           `b
,8888           I8888888I        d8888888888I          ,88888888b           8,
I8888           88888888b       d88888888888'          8888888888b          8I
d8886           888888888       Y888888888P'           Y8888888888,        ,8b
88888b          I88888888b      `Y8888888^             `Y888888888I        d88,
Y88888b         `888888888b,      `""""^                `Y8888888P'       d888I
`888888b         88888888888b,                           `Y8888P^        d88888
 Y888888b       ,8888888888888ba,_          _______        `""^        ,d888888
 I8888888b,    ,888888888888888888ba,_     d88888888b               ,ad8888888I
 `888888888b,  I8888888888888888888888b,    ^"Y888P"^      ____.,ad88888888888I
  88888888888b,`888888888888888888888888b,     ""      ad888888888888888888888'
  8888888888888698888888888888888888888888b_,ad88ba,_,d88888888888888888888888
  88888888888888888888888888888888888888888b,`"""^ d8888888888888888888888888I
  8888888888888888888888888888888888888888888baaad888888888888888888888888888'
  Y8888888888888888888888888888888888888888888888888888888888888888888888888P
  I888888888888888888888888888888888888888888888P^  ^Y8888888888888888888888'
  `Y88888888888888888P88888888888888888888888888'     ^88888888888888888888I
   `Y8888888888888888 `8888888888888888888888888       8888888888888888888P'
    `Y888888888888888  `888888888888888888888888,     ,888888888888888888P'
     `Y88888888888888b  `88888888888888888888888I     I888888888888888888'
       "Y8888888888888b  `8888888888888888888888I     I88888888888888888'
         "Y88888888888P   `888888888888888888888b     d8888888888888888'
            ^""""""""^     `Y88888888888888888888,    888888888888888P'
                             "8888888888888888888b,   Y888888888888P^
                              `Y888888888888888888b   `Y8888888P"^
                                "Y8888888888888888P     `""""^
                                  `"YY88888888888P'
                                       ^""""""""'
*/
#include<stdio.h>
#include<cstring>
#include<algorithm>
#define AC main()
using namespace std;
const int MYDD = 1103;

int AC {
	int n, k, len[MYDD], lenans;
	char a[113];
	scanf("%d %d", &n, &k);

	for(int j = 0; j < MYDD; j++)  len[j] = 0;
	for(int j = 0; j < n + 1; j++) {
		scanf("%s", a);
		if(j == n) {
			lenans = strlen(a);//答案字符串的长度
			break;
		}
		len[strlen(a)]++;

	}

	int cnt = 0;
	for(int j = 0; j < lenans; j++) {
		cnt += len[j];
	}
	int Min = cnt + cnt/k * 5 + 1;

	cnt = -1;
	for(int j = 0; j <= lenans; j++) {
		cnt += len[j];
	}
	int Max = cnt + cnt/k * 5 + 1;
	printf("%d %d\n", Min, Max);
	return 0;
}

/*
5 4
12
23
43
546
123
123

4 10



*/