#include <bits/stdc++.h>
using namespace std;
int n, t;
string s1, s2, s3;
int main() {
  cin >> n >> t >> s1 >> s2;
  int counter1 = 0, counter2 = 0, nreq = 0;
  for (int i = 0; i < n; i++) s3 += " ";
  if (t == 0 && s1 != s2)
    printf("-1");
  else {
    if (t == 0) {
      cout << s1;
      return 0;
    }
    for (int i = 0; i < n; i++) nreq += (s1[i] == s2[i]);
    for (int i = 0; i < n; i++)
      if (s1[i] == s2[i] && nreq && counter1 < n - t && counter2 < n - t)
        s3[i] = s1[i], counter1++, counter2++, nreq--;
    for (int i = 0; i < n; i++)
      if (s3[i] == ' ')
        if (s1[i] != s2[i] && counter1 < n - t)
          counter1++, s3[i] = s1[i];
        else if (s1[i] != s2[i] && counter2 < n - t)
          counter2++, s3[i] = s2[i];
        else if (s1[i] == s2[i])
          s3[i] = 'a' + (s1[i] == 'a');
        else
          s3[i] =
              'a' + (min(s1[i], s2[i]) == 'a') + (s1[i] + s2[i] == 'a' + 'b');
    cout << ((counter1 < n - t || counter2 < n - t) ? "-1" : s3);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 n, t;
string s1, s2, s3;
int main() {
  cin >> n >> t >> s1 >> s2;
  int counter1 = 0, counter2 = 0, nreq = 0;
  for (int i = 0; i < n; i++) s3 += " ";
  if (t == 0 && s1 != s2)
    printf("-1");
  else {
    if (t == 0) {
      cout << s1;
      return 0;
    }
    for (int i = 0; i < n; i++) nreq += (s1[i] == s2[i]);
    for (int i = 0; i < n; i++)
      if (s1[i] == s2[i] && nreq && counter1 < n - t && counter2 < n - t)
        s3[i] = s1[i], counter1++, counter2++, nreq--;
    for (int i = 0; i < n; i++)
      if (s3[i] == ' ')
        if (s1[i] != s2[i] && counter1 < n - t)
          counter1++, s3[i] = s1[i];
        else if (s1[i] != s2[i] && counter2 < n - t)
          counter2++, s3[i] = s2[i];
        else if (s1[i] == s2[i])
          s3[i] = 'a' + (s1[i] == 'a');
        else
          s3[i] =
              'a' + (min(s1[i], s2[i]) == 'a') + (s1[i] + s2[i] == 'a' + 'b');
    cout << ((counter1 < n - t || counter2 < n - t) ? "-1" : s3);
  }
  return 0;
}

```