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 = [int(i) for i in input().split()]

s1 = input()
s2 = input()

same = 0

for i in range(n):

    if s1[i] is s2[i]:
        same += 1

diff = n - same

if t * 2 < diff:
    print(-1)
    exit()

if t * 2 == diff:

    f1 = t
    f2 = t

    s3 = ""

    for i in range(n):

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

            s3 = s3 + s1[i]

        else:

            if f1 > 0:

                s3 = s3 + s1[i]
                f1 -= 1

            else:

                s3 = s3 + s2[i]

    print(s3)
    exit()

oth = max(0, t - diff)

f1 = max(0, diff - t)
f2 = max(0, diff - t)

s3 = ""

for i in range(n):

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

        if oth > 0:

            if s1[i] == 'a':

                s3 = s3 + 'b'

            else:

                s3 = s3 + 'a'

            oth -= 1

        else:

            s3 += s1[i]

    else:

        if f1 > 0:

            s3 += s2[i]
            f1 -= 1

        elif f2 > 0:

            s3 += s1[i]
            f2 -= 1


        else:

            if 'a' not in [s1[i], s2[i]]:

                s3 = s3 + 'a'

            elif 'b' not in [s1[i], s2[i]]:

                s3 = s3 + 'b'

            else:

                s3 = s3 + 'c'

# oth = t - f1
# f = 0
#
# if diff % 2 is 1:
#     f = 1
#     oth -= 1
#
# for i in range(n):
#
#     if s1[i] == s2[i]:
#
#         if oth > 0:
#
#             if s1[i] == 'a':
#
#                 s3 = s3 + 'b'
#
#             else:
#
#                 s3 = s3 + 'a'
#
#             oth -= 1
#
#         else:
#             # print(s1)
#             s3 = s3 + s1[i]
#
#     else:
#
#         if f is 1:
#
#             if 'a' not in [s1[i], s2[i]]:
#
#                 s3 = s3 + 'a'
#
#             elif 'b' not in [s1[i], s2[i]]:
#
#                 s3 = s3 + 'b'
#
#             else:
#
#                 s3 = s3 + 'c'
#
#             f = 0
#
#         else:
#
#             if s1[i] == s2[i]:
#
#                 s3 = s3 + s1[i]
#
#             else:
#
#                 if f1 > 0:
#
#                     s3 = s3 + s1[i]
#                     f1 -= 1
#
#                 else:
#
#                     s3 = s3 + s2[i]

print(s3)