Problem: 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.

Solution (in JAVA):

Solution:
import java.io.* ;
import java.util.* ;
public class Codeshefcode{
	public static void main(String[] args) throws IOException{
		Solver Machine = new Solver() ;
		Machine.Solve() ;
		Machine.Finish() ;		
	}	
}
class Mod{
	static long mod=1000000007 ;
	static long d(long a,long b){ return (a*ModInverse(b))%mod ; }
	static long m(long a , long b){ return (a*b)%mod ; }
	static private long ModInverse(long a ){ return pow(a,mod-2) ; }
	static long pow(long a,long b){
		long val=a ; long ans=1 ;
		while(b!=0){
			if((b&1)==1) ans = m(ans,val) ; 
			val = m(val,val) ; 
			b>>=1 ;
		}
		return ans ;
	}	
}
class pair implements Comparable<pair>{
	long x ; int y ;
	pair(long a,int b){ x=a ; y=b ; }
	public int compareTo(pair p){
		return (this.x<p.x ? -1 : (this.x>p.x ? 1 : (this.y<p.y ? -1 : (this.y>p.y ? 1 : 0)))) ; 
	}
}
class Solver{
	FasterScanner ip = new FasterScanner(System.in) ;	  
	PrintWriter op = new PrintWriter(System.out) ;
	void Solve() throws IOException{
		int n = ip.i() ;
		int ct1[] = new int[10] ;
		int ct2[] = new int[10] ;
		char c1[] = ip.S().toCharArray() ;
		char c2[] = ip.S().toCharArray() ;
		for(int i=0 ; i<n ; i++){
			ct1[c1[i]-'0']++ ;
			ct2[c2[i]-'0']++ ;
		}
		op.println(SolveMin(ct1.clone(),ct2.clone())) ;
		op.println(SolveMax(ct1,ct2)) ;
  	}	
	int SolveMax(int a1[],int a2[]){
		int ans=0 ;
		for(int i=9 ; i>=0 ; i--){
			int ptr=i-1 ;
			while(a2[i]!=0 && ptr>=0){
				if(a1[ptr]>=a2[i]){
					ans+=a2[i] ;
					a1[ptr]-=a2[i] ;
					a2[i]=0 ;
				}else{
					ans+=a1[ptr] ;
					a2[i]-=a1[ptr] ;
					a1[ptr]=0 ;
				}
				ptr-- ;
			}
		}		
		return ans ;
	}
	int SolveMin(int a1[],int a2[]){
		for(int i=0 ; i<10 ; i++){
			if(a1[i]>=a2[i]){
				a1[i]-=a2[i] ;
				a2[i]=0 ;
			}else{
				a2[i]-=a1[i] ;
				a1[i]=0 ;
			}
		}
		int ptr1=0 ; int ptr2 ; 
		while(ptr1<10){
			if(a1[ptr1]>0){
				ptr2=ptr1 ;
				while(ptr2<10 && a1[ptr1]!=0){
					if(a1[ptr1]>=a2[ptr2]){
						a1[ptr1]-=a2[ptr2] ;
						a2[ptr2]=0 ;
					}else{
						a2[ptr2]-=a1[ptr1] ;
						a1[ptr1]=0 ;
					}
					ptr2++ ;
				}
			}
			ptr1++ ;
		}
		int ans=0 ;
		for(int i=0 ; i<10 ; i++) ans+=a1[i] ;
		return ans ;
	}
	void Finish(){ 
		op.flush();
		op.close();
	}	
}
@SuppressWarnings("serial")
class mylist extends ArrayList<pair>{
	
}
@SuppressWarnings("serial")
class MySet extends TreeSet<pair>{
	
}
class FasterScanner { 
	private InputStream mIs;
	private byte[] buf = new byte[1024];
	private int curChar,numChars;
	public FasterScanner() { this(System.in); } 
	public FasterScanner(InputStream is) { mIs = is;} 
	public int read() {
		if (numChars == -1) throw new InputMismatchException();    
		if (curChar >= numChars) {
			curChar = 0;
			try { numChars = mIs.read(buf);} catch (IOException e) { throw new InputMismatchException();}
			if (numChars <= 0) return -1;            
		}
		return buf[curChar++];
	} 
	public String nextLine(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isEndOfLine(c));
		return res.toString();
	} 
	public String S(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		}while (!isSpaceChar(c));
		return res.toString();
	} 
	public long l(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		int sgn = 1;
		if (c == '-') { sgn = -1 ; c = read() ; }
		long res = 0;
		do{
			if (c < '0' || c > '9') throw new InputMismatchException();
			res *= 10 ; res += c - '0' ; c = read();
		}while(!isSpaceChar(c));
		return res * sgn;
	} 
	public int i(){
		int c = read() ;
		while (isSpaceChar(c)) c = read();        
		int sgn = 1;
		if (c == '-') { sgn = -1 ; c = read() ; }
		int res = 0;
		do{
			if (c < '0' || c > '9') throw new InputMismatchException();
		res *= 10 ; res += c - '0' ; c = read() ;
		}while(!isSpaceChar(c));
		return res * sgn;
	} 
	public boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } 
	public boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } 
}