#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, t;
  cin >> n >> t;
  string s1, s2, fin(n, 'a');
  cin >> s1 >> s2;
  int simi = n - t, simi2 = simi;
  for (int i = 0; i < n; i++)
    for (char j = 'a'; j <= 'z'; j++)
      if (j != s1[i] && j != s2[i]) {
        fin[i] = j;
        break;
      }
  for (int i = 0; i < n; i++)
    if (s1[i] == s2[i] && simi > 0) fin[i] = s1[i], simi--, simi2--;
  for (int i = 0; i < n; i++)
    if (s1[i] != s2[i] && (simi > 0 || simi2 > 0))
      if (simi > simi2)
        fin[i] = s1[i], simi--;
      else
        fin[i] = s2[i], simi2--;
  if (simi != 0 || simi2 != 0)
    cout << -1 << endl;
  else
    cout << fin << endl;
}

### Prompt

Please formulate a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, t;
  cin >> n >> t;
  string s1, s2, fin(n, 'a');
  cin >> s1 >> s2;
  int simi = n - t, simi2 = simi;
  for (int i = 0; i < n; i++)
    for (char j = 'a'; j <= 'z'; j++)
      if (j != s1[i] && j != s2[i]) {
        fin[i] = j;
        break;
      }
  for (int i = 0; i < n; i++)
    if (s1[i] == s2[i] && simi > 0) fin[i] = s1[i], simi--, simi2--;
  for (int i = 0; i < n; i++)
    if (s1[i] != s2[i] && (simi > 0 || simi2 > 0))
      if (simi > simi2)
        fin[i] = s1[i], simi--;
      else
        fin[i] = s2[i], simi2--;
  if (simi != 0 || simi2 != 0)
    cout << -1 << endl;
  else
    cout << fin << endl;
}

```