#include <bits/stdc++.h>
using namespace std;
char getdif(char a, char b) {
  if (a != 'a' && b != 'a') {
    return 'a';
  }
  if (a != 'b' && b != 'b') {
    return 'b';
  } else
    return 'c';
}
int main() {
  int n, t, c1 = 0, c2 = 0, l, f = 0, i;
  cin >> n >> t;
  char s1[n + 5], s2[n + 5], s3[n + 5];
  cin >> s1;
  cin >> s2;
  for (i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      c1++;
    }
  }
  c2 = n - c1;
  if (t < (c2 + 1) / 2) {
    printf("-1");
    return 0;
  }
  for (i = 0; i <= n; i++) {
    s3[i] = s1[i];
  }
  if (t > c2) {
    for (i = 0; i < n; i++) {
      if (s1[i] != s2[i]) {
        s3[i] = getdif(s1[i], s2[i]);
        t--;
      }
    }
    for (i = 0; i < n; i++) {
      if (s1[i] == s2[i] && t) {
        s3[i] = getdif(s1[i], s2[i]);
        t--;
      }
    }
  } else {
    t = c2 - t;
    t = 2 * t;
    for (i = 0; i < n; i++) {
      if (s1[i] != s2[i]) {
        if (t) {
          if (f == 0) {
            s3[i] = s1[i];
            f = 1;
          } else {
            s3[i] = s2[i];
            f = 0;
          }
          t--;
        } else
          s3[i] = getdif(s1[i], s2[i]);
      }
    }
  }
  cout << s3;
  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>
using namespace std;
char getdif(char a, char b) {
  if (a != 'a' && b != 'a') {
    return 'a';
  }
  if (a != 'b' && b != 'b') {
    return 'b';
  } else
    return 'c';
}
int main() {
  int n, t, c1 = 0, c2 = 0, l, f = 0, i;
  cin >> n >> t;
  char s1[n + 5], s2[n + 5], s3[n + 5];
  cin >> s1;
  cin >> s2;
  for (i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      c1++;
    }
  }
  c2 = n - c1;
  if (t < (c2 + 1) / 2) {
    printf("-1");
    return 0;
  }
  for (i = 0; i <= n; i++) {
    s3[i] = s1[i];
  }
  if (t > c2) {
    for (i = 0; i < n; i++) {
      if (s1[i] != s2[i]) {
        s3[i] = getdif(s1[i], s2[i]);
        t--;
      }
    }
    for (i = 0; i < n; i++) {
      if (s1[i] == s2[i] && t) {
        s3[i] = getdif(s1[i], s2[i]);
        t--;
      }
    }
  } else {
    t = c2 - t;
    t = 2 * t;
    for (i = 0; i < n; i++) {
      if (s1[i] != s2[i]) {
        if (t) {
          if (f == 0) {
            s3[i] = s1[i];
            f = 1;
          } else {
            s3[i] = s2[i];
            f = 0;
          }
          t--;
        } else
          s3[i] = getdif(s1[i], s2[i]);
      }
    }
  }
  cout << s3;
  return 0;
}

```