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:
n, t = map(int, input().split())

s1 = input()

s2 = input()

ans = ['' for i in range(n)]

k = n - t

for i in range(n):

    if k == 0:

        break

    if s1[i] == s2[i]:

        ans[i] = s1[i]

        k -= 1

k1 = k

k2 = k

for i in range(n):

    if ans[i] != '': continue

    if 0 != k1 >= k2:

        ans[i] = s1[i]

        k1 -= 1

    elif 0 != k2 >= k1:

        ans[i] = s2[i]

        k2 -= 1

    elif k1 == k2 == 0:

        if s1[i] != 'a' != s2[i]:

            ans[i] = 'a'

        elif s1[i] != 'b' != s2[i]:

            ans[i] = 'b'

        else:

            ans[i] = 'c'

if k1 > 0 or k2 > 0:

    print(-1)

else:

    print(*ans, sep='')

    



# Made By Mostafa_Khaled