#include <bits/stdc++.h>
using namespace std;
int n, cnt;
char s[100005], t[100005], ans[100005];
char getC(char a, char b) {
  if (a != 'a' && b != 'a') return 'a';
  if (a != 'b' && b != 'b') return 'b';
  return 'c';
}
int main() {
  scanf("%d%d", &n, &cnt);
  scanf("%s%s", s, t);
  int dif = 0;
  for (int i = 0; i < (int)(n); ++i) dif += s[i] != t[i];
  bool flag = false;
  for (int i = 0; i < (int)(n); ++i)
    if (s[i] != t[i]) {
      if (dif == 1 && !flag) {
        ans[i] = getC(s[i], t[i]);
        --cnt;
      } else if (flag) {
        ans[i] = s[i];
        --cnt;
      } else {
        ans[i] = t[i];
      }
      flag = !flag;
      --dif;
    }
  if (cnt < 0) {
    printf("-1\n");
    return 0;
  }
  for (int i = 0; i < (int)(n); ++i)
    if (s[i] == t[i]) {
      if (cnt == 0) {
        ans[i] = s[i];
      } else {
        ans[i] = (s[i] == 'a') ? 'b' : 'a';
        --cnt;
      }
    }
  cnt *= 2;
  for (int i = 0; i < (int)(n); ++i)
    if (s[i] != t[i]) {
      if (ans[i] != s[i] && ans[i] != t[i]) continue;
      if (cnt == 0) continue;
      ans[i] = getC(s[i], t[i]);
      --cnt;
    }
  ans[n] = '\0';
  printf("%s\n", ans);
  return 0;
}

### Prompt

Generate 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 n, cnt;
char s[100005], t[100005], ans[100005];
char getC(char a, char b) {
  if (a != 'a' && b != 'a') return 'a';
  if (a != 'b' && b != 'b') return 'b';
  return 'c';
}
int main() {
  scanf("%d%d", &n, &cnt);
  scanf("%s%s", s, t);
  int dif = 0;
  for (int i = 0; i < (int)(n); ++i) dif += s[i] != t[i];
  bool flag = false;
  for (int i = 0; i < (int)(n); ++i)
    if (s[i] != t[i]) {
      if (dif == 1 && !flag) {
        ans[i] = getC(s[i], t[i]);
        --cnt;
      } else if (flag) {
        ans[i] = s[i];
        --cnt;
      } else {
        ans[i] = t[i];
      }
      flag = !flag;
      --dif;
    }
  if (cnt < 0) {
    printf("-1\n");
    return 0;
  }
  for (int i = 0; i < (int)(n); ++i)
    if (s[i] == t[i]) {
      if (cnt == 0) {
        ans[i] = s[i];
      } else {
        ans[i] = (s[i] == 'a') ? 'b' : 'a';
        --cnt;
      }
    }
  cnt *= 2;
  for (int i = 0; i < (int)(n); ++i)
    if (s[i] != t[i]) {
      if (ans[i] != s[i] && ans[i] != t[i]) continue;
      if (cnt == 0) continue;
      ans[i] = getC(s[i], t[i]);
      --cnt;
    }
  ans[n] = '\0';
  printf("%s\n", ans);
  return 0;
}

```