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:
//package cf;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import java.util.TreeSet;

public class C {

    public static void main(String[] args) {

        Scanner s = new Scanner(System.in);

        int n = s.nextInt();
        int t = s.nextInt();
        s.nextLine();
        String a = s.nextLine();
        String b = s.nextLine();

        char[] kek = new char[]{'a','b','c'};
        
        int countA = n;
        int countB = n;
        
        ArrayList<Integer> eriKohdat = new ArrayList<>();
        TreeSet<Integer> samatKohdat = new TreeSet<>();
        char[] ans = new char[n];
        for (int i = 0; i < n; i++) {
            char aa = a.charAt(i);
            char bb = b.charAt(i);
            if (aa == bb) {
                ans[i] = aa;
                countA--;
                countB--;
                samatKohdat.add(i);
            }
            else {
                eriKohdat.add(i);
                if (kek[0] != aa && kek[0] != bb) ans[i] = kek[0];
                else if (kek[1] != aa && kek[1] != bb) ans[i] = kek[1];
                else ans[i] = kek[2];
            }
        }
        
        //System.out.println(countA + " " + countB);
       
                
        for (Integer i : eriKohdat) {
            if (countA == t && countB == t) {
                for (char an : ans) {
                    System.out.print(an);
                }
                return;
            }
            if (countA == t && countB > t) {
                ans[i] = b.charAt(i);
                countB--;
            } else if (countA > t) {
                ans[i] = a.charAt(i);
                countA--;
            }
            
        }
        while (countA < t) {
            int j = samatKohdat.pollFirst();
            if (ans[j] != 'a') ans[j] = 'a';
            else ans[j] = 'b';
            countA++;
        }
        countA = 0;
        countB = 0;
        for (int i=0; i<ans.length; i++) {
            if (ans[i] != a.charAt(i)) countA++;
            if (ans[i] != b.charAt(i)) countB++;
        }
        if (countA == t && countB == t) {
            for (char an : ans) {
                System.out.print(an);
            }
        }else
            System.out.println("-1");

    }

}