#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, t, di;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  string c;
  t = n - t;
  int cn = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) cn++;
  }
  di = t - cn;
  if (n - cn < 2 * di) {
    cout << -1;
    return 0;
  }
  int ch = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      if (t) {
        c += a[i];
        t--;
      } else {
        char df = 'a';
        while (df == a[i] || df == b[i]) df++;
        c += df;
      }
    } else if (di > 0) {
      if (ch == 0)
        c += a[i];
      else {
        c += b[i];
        di--;
      }
      ch ^= 1;
    } else {
      char df = 'a';
      while (df == a[i] || df == b[i]) df++;
      c += df;
    }
  }
  cout << c << endl;
  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;
int main() {
  int n, t, di;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  string c;
  t = n - t;
  int cn = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) cn++;
  }
  di = t - cn;
  if (n - cn < 2 * di) {
    cout << -1;
    return 0;
  }
  int ch = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      if (t) {
        c += a[i];
        t--;
      } else {
        char df = 'a';
        while (df == a[i] || df == b[i]) df++;
        c += df;
      }
    } else if (di > 0) {
      if (ch == 0)
        c += a[i];
      else {
        c += b[i];
        di--;
      }
      ch ^= 1;
    } else {
      char df = 'a';
      while (df == a[i] || df == b[i]) df++;
      c += df;
    }
  }
  cout << c << endl;
  return 0;
}

```