Codeforces 777B：Game of Credit Cards（貪心）

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.

題意

Sherlock和Moriarty有\(n\)張卡片，每一個卡片上有一個數字，如今有Sherlock和Moriarty 兩我的在比較這些卡片上的數字大小，小的數字須要接受懲罰，Sherlock的卡片順序是固定的，Moriarty的卡片順序能夠隨意變更，求Moriarty的最小接受懲罰次數是多少，Sherlock最大懲罰對方的次數是多少

思路

將兩個字符串轉換成數組，排序比較就好了

求Moriarty的最小接受懲罰次數的時候，能夠反着求：求Moriarty的卡片數字不小於Sherlock的張數，而後用\(n\)減去便可

代碼

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
int s[maxn],m[maxn];
int nums[100],numm[100];
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    string s1,s2;
    cin>>n;
    cin>>s1>>s2;
    for(int i=0;i<n;i++)
        s[i]=s1[i]-'0',m[i]=s2[i]-'0';
    sort(s,s+n);
    sort(m,m+n);
    int pos1=0;
    int pos2=0;
    int res1=0;
    int res2=0;
    for(int i=0;i<n;i++)
    {
        if(pos1>=n&&pos2>=n)
            break;
        while(pos1<n&&m[pos1]<s[i])
            pos1++;
        while(pos2<n&&m[pos2]<=s[i])
            pos2++;
        if(pos1<n)
            res1++,pos1++;
        if(pos2<n)
            res2++,pos2++;
    }
    cout<<n-res1<<endl;
    cout<<res2<<endl;
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
    #endif
    return 0;
}