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 CODEFORCES_777_B {
	public static void main (String [] t) throws IOException {
		INPUT in = new INPUT (System.in);
		PrintWriter out = new PrintWriter (System.out);

		int N = in.iscan ();
		int [] arr = new int [N];
		int [] arr2 = new int [N];
		
		int i = 0;
		
		for (char c : in.sscan ().toCharArray ())
			arr [i++] = Character.getNumericValue (c);
		
		i = 0;
		
		for (char c : in.sscan ().toCharArray ())
			arr2 [i++] = Character.getNumericValue (c);

		int [] arr3 = new int [10];
		
		for (int n = 0; n < N; ++n)
			++arr3 [arr2 [n]];
		
		int min = 0, max = 0;
		int [] aux = Arrays.copyOf (arr3, 10);
		
		// min
		boolean good = false;
		
		for (int n = 0; n < N; ++n) {
			good = false;
			
			for (int j = arr [n]; j <= 9; ++j) {
				if (aux [j] > 0) {
					--aux [j];
					good = true;
					break;
				}
			}
			
			if (!good) {
				++min;
				
				for (int j = 0; j < arr [n]; ++j)
					if (aux [j] > 0) {
						--aux [j];
						break;
					}
			}
		}
		
		aux = Arrays.copyOf (arr3, 10);
		
		// max
		for (int n = 0; n < N; ++n) {
			good = false;
			
			for (int j = arr [n] + 1; j <= 9; ++j) {
				if (aux [j] > 0) {
					--aux [j];
					++max;
					good = true;
					break;
				}
			}
			
			if (!good) {
				for (int j = 0; j <= arr [n]; ++j)
					if (aux [j] > 0) {
						--aux [j];
						break;
					}
			}
		}
		
		out.println (min);
		out.println (max);
		out.close ();
	}

	private static class INPUT {
		
		private InputStream stream;
		private byte [] buf = new byte [1024];
		private int curChar, numChars;
		
		public INPUT (InputStream stream) {
			this.stream = stream;
		}
		
		public INPUT (String file) throws IOException {
			this.stream = new FileInputStream (file);
		}

		public int cscan () throws IOException {
			if (curChar >= numChars) {
				curChar = 0;
				numChars = stream.read (buf);
			}
			
			return buf [curChar++];
		}
		
		public int iscan () throws IOException {
			int c = cscan (), sgn = 1;
			while (space (c)) c = cscan ();
			
			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}
			
			int res = 0;
			
			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));
			
			return res * sgn;
		}
		
		public String sscan () throws IOException {
			int c = cscan ();
			while (space (c)) c = cscan();
			
			StringBuilder res = new StringBuilder();
			
			do
			{
				res.appendCodePoint (c);
				c = cscan ();
			}
			while (!space (c));
			
			return res.toString ();
		}
		
		public double dscan () throws IOException {
			int c = cscan (), sgn = 1;
			while (space (c)) c = cscan ();
			
			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}
			
			double res = 0;
			
			while (!space (c) && c != '.') {
				if (c == 'e' || c == 'E') return res * UTILITIES.fast_pow (10, iscan ());
				res *= 10;
				res += c - '0';
				c = cscan ();
			}
			
			if (c == '.') {
				c = cscan ();
				double m = 1;
				
				while (!space (c)) {
					if (c == 'e' || c == 'E') return res * UTILITIES.fast_pow (10, iscan ());
					
					m /= 10;
					res += (c - '0') * m;
					c = cscan ();
				}
			}
			
			return res * sgn;
		}
		
		public long lscan () throws IOException {
			int c = cscan (), sgn = 1;
			while (space (c)) c = cscan ();
			
			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}
			
			long res = 0;
			
			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
				
			}
			while (!space (c));
			
			return res * sgn;
		}
		
		public boolean space (int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
	}
	
	public static class UTILITIES {
		
		static final double EPS = 10e-6;
		
		public static int lower_bound (int [] arr, int x) {
			int low = 0, high = arr.length, mid = -1;
			
			while (low < high) {
				mid = (low + high) / 2;
				
				if (arr [mid] >= x)
					high = mid;
				else
					low = mid + 1;
			}
			
			return low;
		}
		
		public static int upper_bound (int [] arr, int x) {
			int low = 0, high = arr.length, mid = -1;
			
			while (low < high) {
				mid = (low + high) / 2;
				
				if (arr [mid] > x)
					high = mid;
				else
					low = mid + 1;
			}
			
			return low;
		}
		
		public static int gcd (int a, int b) {
			return b == 0 ? a : gcd (b, a % b);
		}
		
		public static int lcm (int a, int b) {
			return a * b / gcd (a, b);
		}

		public static int fast_pow_mod (int b, int x, int mod) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;
			
			return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
		}

		public static int fast_pow (int b, int x) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow (b * b, x / 2);
			
			return b * fast_pow (b * b, x / 2);
		}
		
		public static long choose (long n, long k) {
			k = Math.min (k, n - k);
			long val = 1;
			
			for (int i = 0; i < k; ++i)
				val = val * (n - i) / (i + 1);
				
			return val;
		}
		
		public static long permute (int n, int k) {
			if (n < k) return 0;
			long val = 1;
			
			for (int i = 0; i < k; ++i)
				val = (val * (n - i));
				
			return val;
		}
	}
}