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().strip()
s2 = input().strip()

a2z = 'abcdefghijklmnopqrstuvwxyz'

def solve(n, t, s1, s2):
    same = 0
    for c1, c2 in zip(s1, s2):
        if c1 == c2:
            same += 1
    dif = n - same
    c_upper = min(dif // 2, t, n - t)
    c_lower = max(0, n - t - same)
    if c_lower > c_upper:
        return -1
    c = c_lower
    c00 = n - t - c
    c11 = t - c
    c01 = c
    c10 = c
    s3 = []
    for c1, c2 in zip(s1, s2):
        if c1 == c2:
            if c00:
                c00 -= 1
                s3.append(c1)
            else:
                c11 -= 1
                s3.append(difchar(c1, c2))
        else:
            if c01:
                c01 -= 1
                s3.append(c2)
            elif c10:
                c10 -= 1
                s3.append(c1)
            else:
                c11 -= 1
                s3.append(difchar(c1, c2))
    return ''.join(s3)

def difchar(c1, c2):
    for c3 in a2z:
        if c3 != c1 and c3 != c2:
            return c3

print(solve(n, t, s1, s2))