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()
same = n - t
ans = ["" for i in range(n)]
for i in range(n):
    if(same==0):
        break
    if s1[i] == s2[i]:
        same -= 1
        ans[i] = s1[i]

same1 = same2 = same
for i in range(n):
    if(ans[i]!=""):
        continue
    if same1 != 0 and same1>=same2:
        ans[i] = s1[i]
        same1 -= 1
    elif same2 != 0 and same2>=same1:
        ans[i] = s2[i]
        same2 -= 1
    elif same1 == 0 and same2 == 0:
        if s1[i] != 'a' and s2[i] != 'a':
            ans[i] = 'a'
        elif s1[i] != 'b' and s2[i] != 'b':
            ans[i] = 'b'
        elif s1[i] != 'c' and s2[i] != 'c':
            ans[i] = 'c'

if same1 > 0 or same2 > 0:
    print(-1)
else:
    print("".join(ans))