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 MarinaAndVasya {
    
    static boolean valid(char[] ans, char[] s, int x) {
        int cnt = 0;
        for (int i = 0; i < ans.length; i++) {
            if (ans[i] != s[i]) cnt++;
        }
        return cnt == x;
    }
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        
        int n = in.nextInt();
        int x = in.nextInt();
        String s1 = in.next();
        String s2 = in.next();
        int t = n - x;
        
        char[] ans = new char[n];
        boolean[] seen = new boolean[n];
        for (int i = 0; i < n && t > 0; i++) {
            if (s1.charAt(i) == s2.charAt(i)) {
                --t;
                ans[i] = s1.charAt(i);
                seen[i] = true;
            }
        }
        
        int turn = 0;
        for (int i = 0; i < n; i++) {
            if (seen[i]) continue;
            if (t > 0) {
                ans[i] = (turn == 0 ? s1 : s2).charAt(i);
                turn = 1 - turn;
                if (turn == 0) --t;
            } else {
                for (char c = 'a'; c <= 'z'; c++) {
                    if (c != s1.charAt(i) && c != s2.charAt(i)) {
                        ans[i] = c;
                        break;
                    }
                }
            }
            
        }
        if (!valid(ans, s1.toCharArray(), x) || !valid(ans, s2.toCharArray(), x)) {
            System.out.println(-1);
        } else {
            System.out.println(String.valueOf(ans));
        }
        
        in.close();
        System.exit(0);
    }
}