#include <bits/stdc++.h>
char a[100010], b[100010], ans[100010];
char dif(char x, char y) {
  if (x != 'a' && y != 'a') return 'a';
  if (x != 'b' && y != 'b') return 'b';
  return 'c';
}
int main() {
  int n, t, cnt = 0;
  scanf("%d %d", &n, &t);
  t = n - t;
  scanf("%s %s", a, b);
  for (int i = 0; i < n; ++i) {
    if (a[i] == b[i]) cnt++;
  }
  if (cnt + n < 2 * t) {
    printf("-1\n");
    return 0;
  }
  int as, bs, als;
  if (cnt > t) {
    as = bs = 0;
    als = t;
  } else {
    as = bs = t - cnt;
    als = cnt;
  }
  for (int i = 0; i < n; ++i) {
    if (a[i] == b[i]) {
      if (als) {
        ans[i] = a[i];
        als--;
      } else {
        ans[i] = dif(a[i], b[i]);
      }
    } else {
      if (as) {
        ans[i] = a[i];
        as--;
      } else if (bs) {
        ans[i] = b[i];
        bs--;
      } else {
        ans[i] = dif(a[i], b[i]);
      }
    }
  }
  ans[n] = '\0';
  printf("%s\n", ans);
}

### Prompt

Create a solution in Cpp for 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>
char a[100010], b[100010], ans[100010];
char dif(char x, char y) {
  if (x != 'a' && y != 'a') return 'a';
  if (x != 'b' && y != 'b') return 'b';
  return 'c';
}
int main() {
  int n, t, cnt = 0;
  scanf("%d %d", &n, &t);
  t = n - t;
  scanf("%s %s", a, b);
  for (int i = 0; i < n; ++i) {
    if (a[i] == b[i]) cnt++;
  }
  if (cnt + n < 2 * t) {
    printf("-1\n");
    return 0;
  }
  int as, bs, als;
  if (cnt > t) {
    as = bs = 0;
    als = t;
  } else {
    as = bs = t - cnt;
    als = cnt;
  }
  for (int i = 0; i < n; ++i) {
    if (a[i] == b[i]) {
      if (als) {
        ans[i] = a[i];
        als--;
      } else {
        ans[i] = dif(a[i], b[i]);
      }
    } else {
      if (as) {
        ans[i] = a[i];
        as--;
      } else if (bs) {
        ans[i] = b[i];
        bs--;
      } else {
        ans[i] = dif(a[i], b[i]);
      }
    }
  }
  ans[n] = '\0';
  printf("%s\n", ans);
}

```