Codeforces Round 401 (Div. 2)
B
Game of Credit Cards
Codeforces
1300
原题
← AC →
翻译中…

After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards.

翻译中…

Rules of this game are simple: each player bring his favourite $*n*$-digit credit card. Then both players name the digits written on their cards one by one. If two digits are not equal, then the player, whose digit is smaller gets a flick (knock in the forehead usually made with a forefinger) from the other player. For example, if $*n* = 3$, Sherlock's card is $123$ and Moriarty's card has number $321$, first Sherlock names $1$ and Moriarty names $3$ so Sherlock gets a flick. Then they both digit $2$ so no one gets a flick. Finally, Sherlock names $3$, while Moriarty names $1$ and gets a flick.

翻译中…

Of course, Sherlock will play honestly naming digits one by one in the order they are given, while Moriary, as a true villain, plans to cheat. He is going to name his digits in some other order (however, he is not going to change the overall number of occurences of each digit). For example, in case above Moriarty could name $1$, $2$, $3$ and get no flicks at all, or he can name $2$, $3$ and $1$ to give Sherlock two flicks.

翻译中…

Your goal is to find out the minimum possible number of flicks Moriarty will get (no one likes flicks) and the maximum possible number of flicks Sherlock can get from Moriarty. Note, that these two goals are different and the optimal result may be obtained by using different strategies.

翻译中…

Input

翻译中…

The first line of the input contains a single integer $*n*$ ($1 ≤ *n* ≤ 1000$) — the number of digits in the cards Sherlock and Moriarty are going to use.

翻译中…

The second line contains $*n*$ digits — Sherlock's credit card number.

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The third line contains $*n*$ digits — Moriarty's credit card number.

翻译中…

Output

翻译中…

First print the minimum possible number of flicks Moriarty will get. Then print the maximum possible number of flicks that Sherlock can get from Moriarty.

翻译中…

Note

翻译中…

First sample is elaborated in the problem statement. In the second sample, there is no way Moriarty can avoid getting two flicks.

翻译中…

样例

Input #1
3
123
321
Output #1
0
2
InputCopy OutputCopy
Input #2
2
88
00
Output #2
2
0
InputCopy OutputCopy
题目信息
题号
777B
时间限制
2 s
内存限制
256 MB
标签
数据结构
动态规划
贪心
排序

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。