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())
a = input()
b = input()
m = 0
for i in range(n):
    if a[i] != b[i]:
        m += 1
if n - t > n - m + (m >> 1):
    print(-1)
else:
    ls = [ ]
    if m <= t:
        for i in range(n):
            if a[i] == b[i]:
                if m == t:
                    ls.append(a[i])
                else:
                    ls.append('a' if a[i] != 'a' else 'b')
                    m += 1
            elif a[i] != 'a' and b[i] != 'a':
                ls.append('a')
            elif a[i] != 'b' and b[i] != 'b':
                ls.append('b')
            else:
                ls.append('c')
        ta = tb = 0
    else:
        ta = tb = m - t
        for i in range(n):
            if a[i] == b[i]:
                ls.append(a[i])
            elif ta > 0:
                ls.append(a[i])
                ta -= 1
            elif tb > 0:
                ls.append(b[i])
                tb -= 1
            elif a[i] != 'a' and b[i] != 'a':
                ls.append('a')
            elif a[i] != 'b' and b[i] != 'b':
                ls.append('b')
            else:
                ls.append('c')
    assert ta == tb == 0
    s = "".join(ls)
    for i in range(n):
        if a[i] != s[i]:
            ta += 1
        if b[i] != s[i]:
            tb += 1
    assert ta == tb == t
    print(s)