#include <bits/stdc++.h>
using namespace std;
int c[100004], i, n, t, q, w, nr, r, d, s;
char a[100004], b[100004];
int main() {
  scanf("%d %d\n", &n, &t);
  for (i = 1; i <= n; i++) scanf("%c", &a[i]);
  scanf("\n");
  for (i = 1; i <= n; i++) scanf("%c", &b[i]);
  for (i = 1; i <= n; i++) {
    if (a[i] == b[i]) c[++nr] = i;
  }
  if ((n - nr) <= t) {
    for (i = 1; i <= n; i++) {
      if (a[i] != b[i]) {
        if (a[i] != 'a' && b[i] != 'a')
          printf("a");
        else if (a[i] != 'b' && b[i] != 'b')
          printf("b");
        else
          printf("c");
      } else if (r < (n - t)) {
        printf("%c", a[i]);
        r++;
      } else {
        if (a[i] != 'a')
          printf("a");
        else
          printf("b");
      }
    }
  } else {
    q = n - nr;
    if (q > (2 * t))
      printf("%d", -1);
    else {
      s = (n - nr - t) * 2;
      if (s > q)
        printf("%d", s);
      else {
        for (i = 1; i <= n; i++) {
          if (a[i] != b[i]) {
            if (d < s) {
              if (d % 2 == 0)
                printf("%c", a[i]);
              else
                printf("%c", b[i]);
              d++;
            } else {
              if (a[i] != 'a' && b[i] != 'a')
                printf("a");
              else if (a[i] != 'b' && b[i] != 'b')
                printf("b");
              else
                printf("c");
            }
          } else
            printf("%c", a[i]);
        }
      }
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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 c[100004], i, n, t, q, w, nr, r, d, s;
char a[100004], b[100004];
int main() {
  scanf("%d %d\n", &n, &t);
  for (i = 1; i <= n; i++) scanf("%c", &a[i]);
  scanf("\n");
  for (i = 1; i <= n; i++) scanf("%c", &b[i]);
  for (i = 1; i <= n; i++) {
    if (a[i] == b[i]) c[++nr] = i;
  }
  if ((n - nr) <= t) {
    for (i = 1; i <= n; i++) {
      if (a[i] != b[i]) {
        if (a[i] != 'a' && b[i] != 'a')
          printf("a");
        else if (a[i] != 'b' && b[i] != 'b')
          printf("b");
        else
          printf("c");
      } else if (r < (n - t)) {
        printf("%c", a[i]);
        r++;
      } else {
        if (a[i] != 'a')
          printf("a");
        else
          printf("b");
      }
    }
  } else {
    q = n - nr;
    if (q > (2 * t))
      printf("%d", -1);
    else {
      s = (n - nr - t) * 2;
      if (s > q)
        printf("%d", s);
      else {
        for (i = 1; i <= n; i++) {
          if (a[i] != b[i]) {
            if (d < s) {
              if (d % 2 == 0)
                printf("%c", a[i]);
              else
                printf("%c", b[i]);
              d++;
            } else {
              if (a[i] != 'a' && b[i] != 'a')
                printf("a");
              else if (a[i] != 'b' && b[i] != 'b')
                printf("b");
              else
                printf("c");
            }
          } else
            printf("%c", a[i]);
        }
      }
    }
  }
  return 0;
}

```