#include <bits/stdc++.h>
int n, t;
char li1[100010], li2[100010], ans[100010];
int main() {
  scanf("%d%d", &n, &t);
  scanf(" %s", li1);
  scanf(" %s", li2);
  memset(ans, 0, sizeof ans);
  int i, num = 0;
  for (i = 0; i < n && num < n - t; i++)
    if (li1[i] == li2[i]) num++, ans[i] = li2[i];
  if ((num + ((n - num) / 2)) < (n - t)) {
    printf("-1");
    return 0;
  }
  i = 0;
  int n1 = 0, nn = n - t - num, n2 = 0;
  while (n1 < nn || n2 < nn) {
    if (ans[i] != 0) {
      i++;
      continue;
    }
    if (n1 < nn)
      ans[i] = li1[i], n1++;
    else
      ans[i] = li2[i], n2++;
    i++;
  }
  while (i < n) {
    if (ans[i] != 0) {
      i++;
      continue;
    }
    if (li1[i] == 'a' || li2[i] == 'a') {
      if (li1[i] == 'b' || li2[i] == 'b')
        ans[i] = 'c';
      else
        ans[i] = 'b';
    } else
      ans[i] = 'a';
    i++;
  }
  printf("%s\n", ans);
  return 0;
}

### 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>
int n, t;
char li1[100010], li2[100010], ans[100010];
int main() {
  scanf("%d%d", &n, &t);
  scanf(" %s", li1);
  scanf(" %s", li2);
  memset(ans, 0, sizeof ans);
  int i, num = 0;
  for (i = 0; i < n && num < n - t; i++)
    if (li1[i] == li2[i]) num++, ans[i] = li2[i];
  if ((num + ((n - num) / 2)) < (n - t)) {
    printf("-1");
    return 0;
  }
  i = 0;
  int n1 = 0, nn = n - t - num, n2 = 0;
  while (n1 < nn || n2 < nn) {
    if (ans[i] != 0) {
      i++;
      continue;
    }
    if (n1 < nn)
      ans[i] = li1[i], n1++;
    else
      ans[i] = li2[i], n2++;
    i++;
  }
  while (i < n) {
    if (ans[i] != 0) {
      i++;
      continue;
    }
    if (li1[i] == 'a' || li2[i] == 'a') {
      if (li1[i] == 'b' || li2[i] == 'b')
        ans[i] = 'c';
      else
        ans[i] = 'b';
    } else
      ans[i] = 'a';
    i++;
  }
  printf("%s\n", ans);
  return 0;
}

```