cf 777B _ B. Game of Credit Cards (排序比较)

http://codeforces.com/contest/777/problem/B

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.

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.

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.

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.

Input

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.

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

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

Output

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.

Examples

input

3
123
321

output

0
2

input

2
88
00

output

2
0

Note

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

题目大意：有两个人M和S，这两个人都有一个卡号，上面是一串数字。对他们的每一位数字进行比较，数字小的会得到一个flick,，相等时都不得到，，然而玩家M很坏，他可以变换数字的位置，问M可以得到最少的flicks，和S可以得到最多的flick

题解：以为M可以变换位置，那么主动权就在M手里，一个人的变化，相当于另一个也在变了（想象一下参照物的原理），那么对他们的位置进行排序就行了，在寻找最优时，因为主动权在M,所以满足M的所有要求就可以了。

#include <iostream>
#include <bits/stdc++.h>

using namespace std;

char a[1111];
char b[1111];
bool cmp(char x,char y)
{
return x<y;
}
int main()
{
int n;
while(cin>>n){
cin>>a>>b;
sort(a,a+n,cmp);
sort(b,b+n,cmp);
int ans=0;
int i,j;
i=j=0;
while(i<n&&j<n)
{
if(b[j]>=a[i])
{
i++;
j++;
}
else
{
j++;
ans++;
}
}
i=j=0;
int ant=0;
while(i<n&&j<n)
{
if(b[j]>a[i])
{
i++;
j++;
ant++;
}
else
{
j++;
}
}
cout<<ans<<endl<<ant<<endl;
}
return 0;
}