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.*;

public class B {

	public static void main(String[] args) {
		Scanner s = new Scanner (System.in);
		int n = s.nextInt(), t = s.nextInt();
		t=n-t;
		String s1=s.next(), s2=s.next();
		int[] a = new int [n];
		char [] b = new char [n];
		int k=0;
		for(int i=0; i<n; i++) {
			if(s1.charAt(i)==s2.charAt(i)) {
				a[i]=1;
				k++;
			}
		}
		int ans = 0;
		if(t<=k) {
			int x=0;
			for(int i=0; i<n; i++) {
				if(a[i]==0 || x>=t) {
					if(s1.charAt(i)!='a' && s2.charAt(i)!='a') b[i] = 'a';
					else if(s1.charAt(i)!='b' && s2.charAt(i)!='b') b[i] = 'b';
					else b[i] = 'c';
				}
				if(a[i]==1 && x<t) {
					b[i] = s1.charAt(i);
					x++;
				}
			}
		}
		else {
			if((n-k)>=2*(t-k)) {
				int y=0;
				for(int i=0; i<n; i++) {
					if(a[i]==0){
						if(y<t-k) {
							b[i] = s1.charAt(i);
//							System.out.println(b[i]);
							y++;
						}
						else if(t-k<=y&&y<2*(t-k)){
							b[i] = s2.charAt(i);
							y++;
						}
						else {
							if(s1.charAt(i)!='a' && s2.charAt(i)!='a') b[i] = 'a';
							else if(s1.charAt(i)!='b' && s2.charAt(i)!='b') b[i] = 'b';
							else b[i] = 'c';
						}
					}
					else b[i] = s1.charAt(i);
				}
			}
			else ans =-1;
		}
		if(ans != -1)
			for(int i=0; i<n; i++)
				System.out.print(b[i]);
		else
			System.out.print(-1);
	}
}