BE 題意 www. finally tdi code fir eterm htm

B. Treasure Hunt time limit per test1 second memory limit per test：256 megabytes input：standard input output：standard output

After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.

The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.

A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let‘s call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of

77 because its subribbon a appears $\mathrm{77\; times,\; and\; the\; ribbon\; abcdabc\; has\; the\; beauty\; of\; 22\; because\; its\; subribbon\; abc\; appears\; twice.}$

The rules are simple. The game will have $\mathrm{nn\; turns.\; Every\; turn,\; each\; of\; the\; cats\; must\; change\; strictly\; one\; color\; (at\; one\; position)\; in\; his/her\; ribbon\; to\; an\; arbitrary\; color\; which\; is\; different\; from\; the\; unchanged\; one.\; For\; example,\; a\; ribbon\; aaab\; can\; be\; changed\; into\; acab\; in\; one\; turn.\; The\; one\; having\; the\; most\; beautiful\; ribbon\; after}$

nn turns wins the treasure.

Could you find out who is going to be the winner if they all play optimally?

Input

The first line contains an integer $\mathrm{nn\; (0\le n\le 1090\le n\le 109)\; —\; the\; number\; of\; turns.}$

Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than ${\mathrm{105105\; uppercase\; and\; lowercase\; Latin\; letters\; and\; is\; not\; empty.\; It\; is\; guaranteed\; that\; the\; length\; of\; all\; ribbons\; are\; equal\; for\; the\; purpose\; of\; fairness.\; Note\; that\; uppercase\; and\; lowercase\; letters\; are\; considered\; different\; colors.}}^{}$

Output

Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".

Examples Input

3
Kuroo
Shiro
Katie

Output

Kuro

Input

7
treasurehunt
threefriends
hiCodeforces

Output

Shiro

Input

1
abcabc
cbabac
ababca

Output

Katie

Input

15
foPaErcvJ
mZaxowpbt
mkuOlaHRE

Output

Draw

Note

In the first example, after $\mathrm{33\; turns,\; Kuro\; can\; change\; his\; ribbon\; into\; ooooo,\; which\; has\; the\; beauty\; of\; 55,\; while\; reaching\; such\; beauty\; for\; Shiro\; and\; Katie\; is\; impossible\; (both\; Shiro\; and\; Katie\; can\; reach\; the\; beauty\; of\; at\; most\; 44,\; for\; example\; by\; changing\; Shiro‘s\; ribbon\; into\; SSiSS\; and\; changing\; Katie‘s\; ribbon\; into\; Kaaaa).\; Therefore,\; the\; winner\; is\; Kuro.}$

In the fourth example, since the length of each of the string is $\mathrm{99\; and\; the\; number\; of\; turn\; is\; 1515,\; everyone\; can\; change\; their\; ribbons\; in\; some\; way\; to\; reach\; the\; maximal\; beauty\; of\; 99\; by\; changing\; their\; strings\; into\; zzzzzzzzz\; after\; 9\; turns,\; and\; repeatedly\; change\; their\; strings\; into\; azzzzzzzz\; and\; then\; into\; zzzzzzzzz\; thrice.\; Therefore,\; the\; game\; ends\; in\; a\; draw.}$

$\mathrm{題意：給你一個n和三個字符串，表示操作n次（必須操作），每次必須修改一個字符，問最後三個字符串中，重復次數最多的是哪個。}$

$\mathrm{思路：首先我們遍歷一遍，求出每個字符串中單個字符出現次數最多的一個（由貪心知一個子字符串出現幾次，那麽裏面的元素也會出現那麽多次，要想讓它重復次數最多，肯定是一個字符重復次數越多越好）。如果len=1的話，那麽三者肯定會一樣；如果重復的最多次數等於len，那麽就判斷n，如果n等於1，那麽當前字符最大次數為len-1,否則和最大次數不等於len一樣為min(len,mx+n）}$

代碼實現如下：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int n;
 5 string s[3];
 6 int num[3][128], mx[3];
 7 
 8 int main() {
 9     ios::sync_with_stdio(false);
10     cin.tie(0);
11     cin >>n;
12     int len;
13     memset(num, 0, sizeof(num));
14     for(int i = 0; i < 3; i++) {
15         cin >>s[i];
16         len = s[i].size();
17         for(int v : s[i]) {
18             num[i][v]++;
19         }
20         mx[i] = *max_element(num[i], num[i] + 128);
21         if(mx[i] == len) {
22             if(n == 1) {
23                 mx[i] = len - 1;
24             } else {
25                 mx[i] = min(len, mx[i] + n);
26             }
27         }
28         else {
29             mx[i] = min(len, mx[i] + n);
30         }
31     }
32     if(len == 1) cout <<"Draw" <<endl;
33     else if(count(mx, mx + 3, *max_element(mx, mx + 3)) > 1) cout <<"Draw" <<endl;
34     else {
35         if(mx[0] > max(mx[1], mx[2])) cout <<"Kuro" <<endl;
36         else if(mx[1] > max(mx[0], mx[2])) cout <<"Shiro" <<endl;
37         else cout <<"Katie" <<endl;
38     }
39     return 0;
40 }

Codeforces Round #482 (Div. 2) B題