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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class Main {
	static int n,t;
	static char[] a;
	static char[] b;
	static int a1,a2,b1,c,difs,eqs;
	public static void main(String[] args) throws IOException {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer(br.readLine());
		n=Integer.parseInt(st.nextToken());
		t=Integer.parseInt(st.nextToken());
		a=br.readLine().toCharArray();
		b=br.readLine().toCharArray();
		fillDifs();
		a1=-1;
		for (int i = 0; i <= difs/2; i++) {
			if (i+(difs-2*i)<=t && t-i-(difs-2*i)<=eqs){
				a1=i;
				break;
			}
		}
		if (a1==-1) System.out.println(-1);	
		else{
			a2=a1;
			b1=difs-2*a1;
			c=t-a1-(difs-2*a1);
			char[] ans=new char[n];
			for (int i = 0; i < n; i++) {
				if (a[i]==b[i]){
					if (c>0){
						ans[i]=different(a[i], b[i]);
						c--;
					}
					else{
						ans[i]=a[i];
					}
				}
				else{
					if (b1>0){
						ans[i]=different(a[i], b[i]);
						b1--;
					}
					else if (a1>0){
						ans[i]=b[i];
						a1--;
					}
					else{
						ans[i]=a[i];
						a2--;
					}
				}
			}
			System.out.println(ans);
		}
	}
	private static void fillDifs() {
		difs=0;
		for (int i = 0; i < n; i++) if (a[i]!=b[i]) difs++;
		eqs=n-difs;
	}
	static char different(char c1, char c2){
		if (c1!='a' && c2!='a') return 'a';
		else if (c1!='b' && c2!='b') return 'b';
		return 'c';
	}
}