Problem: Marina loves strings of the same length and Vasya loves when there is a third string, different from them in exactly t characters. Help Vasya find at least one such string.

More formally, you are given two strings s1, s2 of length n and number t. Let's denote as f(a, b) the number of characters in which strings a and b are different. Then your task will be to find any string s3 of length n, such that f(s1, s3) = f(s2, s3) = t. If there is no such string, print  - 1.

Input

The first line contains two integers n and t (1 ≤ n ≤ 105, 0 ≤ t ≤ n).

The second line contains string s1 of length n, consisting of lowercase English letters.

The third line contain string s2 of length n, consisting of lowercase English letters.

Output

Print a string of length n, differing from string s1 and from s2 in exactly t characters. Your string should consist only from lowercase English letters. If such string doesn't exist, print -1.

Examples

Input

3 2
abc
xyc


Output

ayd

Input

1 0
c
b


Output

-1

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {
	public static char abc(char p1, char p2) {
		char temp;
		if (p1 != 'a' && p2 != 'a')
			temp = 'a';
		else {
			if (p1 != 'b' && p2 != 'b')
				temp = 'b';
			else
				temp = 'c';

		}
		return temp;
	}

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			int n = sc.nextInt();
			int t = sc.nextInt();
			int same = n - t, counter = 0;
			String s1 = sc.next();
			String s2 = sc.next();
			char c1[] = s1.toCharArray();
			char c2[] = s2.toCharArray();
			char c3[] = new char[c1.length];
			for (int i = 0; i < s1.length(); i++) {
				if (c1[i] == c2[i]) {
					counter++;
					if (counter <= same)
						c3[i] = c1[i];
					else if (c1[i] != 'a')
						c3[i] = 'a';
					else
						c3[i] = 'b';
				}
			}
			if (2 * same > (c1.length + counter))
				System.out.println(-1);
			else if (counter >= same) {
				for (int i = 0; i < s1.length(); i++) {
					if (c1[i] != c2[i])
						c3[i] = Main.abc(c1[i], c2[i]);
				}
				System.out.println(c3);
			} else {
				int q1 = same - counter, q2 = same - counter;
				for (int i = 0; i < s1.length(); i++)
					if (c1[i] == c2[i])
						continue;
					else if (q1 > 0) {
						c3[i] = c1[i];
						q1--;

					} else {
						if (q2 > 0) {
							c3[i] = c2[i];
							q2--;
						} else
							c3[i] = Main.abc(c1[i], c2[i]);

					}
				System.out.println(c3);

			}
		}
	}
}