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;

/**
 * Created by slycer on 10/6/15.
 */
public class MarinaandVasya {

    public static void main ( String [] args ){
        Scanner s = new Scanner( System.in );
        int n = s.nextInt();
        int delta = s.nextInt();
        s.nextLine();
        String a = s.nextLine();
        String b = s.nextLine();
        //System.out.println ( a + " " + b );
        char [] sol = new char[n];

        int egale = n - delta;

        for ( int i=0; i<n; i++ ){
            if ( egale > 0 && a.charAt(i) == b.charAt(i) ){
                egale--;
                sol[i] = a.charAt(i);
            }
        }

        int left = egale;
        for ( int i=0; i<n; i++ ){
            if ( left > 0 && sol[i] == 0 ){
                left--;
                sol[i] = a.charAt(i);
            }
        }

        int right = egale;
        for ( int i=0; i<n; i++ ){
            if ( right > 0 && sol[i] == 0 ){
                right--;
                sol[i] = b.charAt(i);
            }
        }

        for ( int i=0; i<n; i++ ){
            if ( sol[i] == 0 ){
                sol[i] = different( a.charAt(i), b.charAt(i) );
            }
        }

        // daca am completat toate pozitiile egale
        if ( left == 0 && right == 0 ){
            for ( int i=0; i<n; i++ ){
                System.out.print( sol[i] ) ;
            }
        } else {
            System.out.println( "-1" );
        }

    }

    public static char different( char a, char b ){
        for ( char i='a'; i<='z'; i++ ){
            if ( a != i && b != i ){
                return i;
            }
        }
        return 'a';
    }

}