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:
def f(sa, sb):
    ans = 0
    for i in range(len(sa)):
        ans += sa[i] != sb[i]
    return ans

def third(a0, b0):
    if a0 != "a" and b0 != "a":
        return "a"
    if a0 != "b" and b0 != "b":
        return "b"
    if a0 != "c" and b0 != "c":
        return "c"
n, t = list(map(int, input().split()))
s1 = input()
s2 = input()
sovp = n - f(s1, s2) #совпадающих из нашей строки
needsovp = n - t #необходимо совпадающих
ans = [0] * n
need1r = needsovp - sovp
need2r = needsovp - sovp
if sovp >= needsovp:
    for i in range(n):
        if s1[i] != s2[i]:
            ans[i] = third(s1[i], s2[i])
        elif needsovp > 0:
            ans[i] = s1[i]
            needsovp -= 1
        else:
            ans[i] = third(s1[i], s2[i])
    if not(f("".join(ans), s1) == f("".join(ans), s2) == t):
        print(-1)
    else:
        print(*ans, sep="")
else:
    for i in range(n):
        if s1[i] != s2[i]:
            if need1r > 0:
                need1r -= 1
                ans[i] = s1[i]
            elif need2r > 0:
                need2r -= 1
                ans[i] = s2[i]
            else:
                ans[i] = third(s1[i], s2[i])
        else:
            ans[i] = s1[i]
    if not(f("".join(ans), s1) == f("".join(ans), s2) == t):
        print(-1)
    else:
        print(*ans, sep="")