CF 777B

B. Game of Credit Cards

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 1and 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

3123321

output

02

input

28800

output

20

Note

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

/*  题意类似田忌赛马，输入一个n，A和B数组中都有n个数 然后安排A数组和B数组中数的位置  然后对应匹配，如果ai>bi（i：0~n-1）则b获得一个惩罚，相等则两人都不会获得惩罚。  输出两个答案 B的惩罚最少是多少 A的惩罚最多是多少。  因为这两个事件是独立的 B的惩罚最少不一定满足A的惩罚最多 所以应该分情况讨论  先将A和B数组排序。  让 B 输的尽可能的少：排序后，从B最大的数到最小的数（n-1到0）一一匹配A中的数。  让 A 输的尽可能的多：排序后，从A最小的数到最大的数（0到n-1）一一匹配B中的数。*/#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <string>#include <cstdlib>using namespace std;const int maxn = 1010;int a[maxn];int b[maxn];int main(void){int n;scanf("%d",&n);for(int i=0;i<n;i++)        scanf("%1d",&a[i]);for(int i=0;i<n;i++)        scanf("%1d",&b[i]);    sort(a,a+n);    sort(b,b+n);    int ans=n-1;    for(int i=n-1;i>=0;i--)        if(b[ans]>=a[i]) ans--;    cout<<ans+1<<endl;    ans=0;    for(int i=0;i<n;i++)        if(b[i]>a[ans]) ans++;    cout<<ans<<endl;return 0;}