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 PYTHON3):

Solution:
String = ['x', 'y', 'z']
n, t = map(int, input().split())
s1 = input()
s2 = input()
k = n - t
count = 0
ans = ''
for i in range(n):
    if s1[i] == s2[i]:
        count += 1
number = n - count 
if count >= k:
    for i in range(n):
        if s1[i] == s2[i] and k > 0:
            ans += s1[i]
            k -= 1
        else:
            for s in String:
                if s != s1[i] and s != s2[i]:
                    ans += s
                    break
    print(ans)
elif number > 2 * t:
    print('-1')
elif number <= 2 * t:
    ballance1 = number - t
    ballance2 = number - t
    for i in range(n):
        if s1[i] == s2[i]:
            ans += s1[i]
        elif ballance1:
            ans += s1[i]
            ballance1 -= 1
        elif ballance2:
            ans += s2[i]
            ballance2 -= 1
        else:
            for s in String:
                if s != s1[i] and s != s2[i]:
                    ans += s
                    break
    print(ans)