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 final class C234
{
    static int n;

    static int cntdiff(String a, String b)
    {
        int res = 0;
        for (int i = 0; i < n; ++i)
            if (a.charAt(i) != b.charAt(i)) ++res;
        return res;
    }

    public static void main(String args[])
    {
        Scanner cin = new Scanner(System.in);
        n = cin.nextInt();
        int t = cin.nextInt();
        String a = cin.next();
        String b = cin.next();
        if (t == 0)
        {
            if (a.equals(b)) System.out.print(a);
            else System.out.print(-1);
            return;
        }
        StringBuilder ans = new StringBuilder("");
        int c1 = 0, c2 = 0, l1 = -1;
        for (int i = 0; i < n && (c1 < t || c2 < t); ++i)
        {
            if (a.charAt(i) == b.charAt(i)) {
                ans.append(a.charAt(i));
            }
            else if (c1 == c2) {
                ans.append(a.charAt(i));
                ++c1;
                l1 = i;
            }
            else {
                ans.append(b.charAt(i));
                ++c2;
            }
        }
        if (ans.length() < n)
        {
            for (int i = ans.length(); i < n; ++i) {
                if (a.charAt(i) != b.charAt(i)) {
                    System.out.print(-1);
                    return;
                }
                ans.append(a.charAt(i));
            }
        }
        if (c1 > c2)
        {
            ++c2;
            for (char c = 'a'; c <= 'z'; ++c)
                if (a.charAt(l1) != c && b.charAt(l1) != c)
                {
                    ans.setCharAt(l1, c);
                    break;
                }
        }
        for (int i = 0; i < n && (c1 < t || c2 < t); ++i)
        {
            if (a.charAt(i) == b.charAt(i)) {
                for (char c = 'a'; c <= 'z'; ++c)
                    if (a.charAt(i) != c && b.charAt(i) != c)
                    {
                        ans.setCharAt(i, c);
                        ++c1;
                        ++c2;
                        break;
                    }
            }
        }
        for (int i = 0; i < n && (c1 < t || c2 < t); ++i)
        {
            if (a.charAt(i) == b.charAt(i)) continue;
            if (a.charAt(i) == ans.charAt(i) && c1 == c2)
            {
                ++c1;
                for (char c = 'a'; c <= 'z'; ++c)
                    if (a.charAt(i) != c && b.charAt(i) != c)
                    {
                        ans.setCharAt(i, c);
                        break;
                    }
            }
            else if (b.charAt(i) == ans.charAt(i))
            {
                ++c2;
                for (char c = 'a'; c <= 'z'; ++c)
                    if (a.charAt(i) != c && b.charAt(i) != c)
                    {
                        ans.setCharAt(i, c);
                        break;
                    }
            }
        }
        if (c1 < t || c2 < t) System.out.print(-1);
        else System.out.print(ans);
    }
}