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:
/*input
3 2
abc
xyc
*/

import java.util.*;

public class hello {
	public static void main (String args[]) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int t = sc.nextInt();
		char[] a = sc.next().toCharArray();
		char[] b = sc.next().toCharArray();
		int cnt = 0;
		for(int i = 0; i<n; i++)
			if(a[i] != b[i]) cnt++;
		if(t < n - (n-cnt+cnt/2))
			System.out.println(-1);
		else{
			char[] ans = new char[n];
			for(int i = 0; i<n; i++)
			{
				for(char x = 'a'; x <= 'z'; x++)
					if(x != a[i] && x != b[i]) {
						ans[i] = x;
						break;
					}
			}
			cnt = n;
			for(int i = 0; i<n && t!=cnt; i++)
			{
				if(a[i] == b[i]) {
					cnt--;
					ans[i] = a[i];
				}
			}
			if(t!=cnt){
				int c = 0;
				for(int i = 0; i<n && t!=cnt; i++)
				{
					if(a[i] != b[i]){
						if(c%2==1) {
							cnt--;
							ans[i] = a[i];
						}
						else ans[i] = b[i];
						c++;
					}
				}
			}
			System.out.println(ans);
		}

	}
}