#include <bits/stdc++.h>
using namespace std;
char a[1000000];
int main() {
  int i, j, k, n, m, l1, l2;
  string s, s1;
  cin >> n >> m >> s >> s1;
  l1 = l2 = n - m;
  for (i = 0; i < n; i++) {
    if (s[i] == s1[i] && l1 && l2) {
      a[i] = s[i];
      l1--;
      l2--;
    }
  }
  for (i = 0; i < n; i++) {
    if (!a[i] && l1) {
      a[i] = s[i];
      l1--;
    }
  }
  for (i = 0; i < n; i++) {
    if (!a[i] && l2) {
      a[i] = s1[i];
      l2--;
    }
  }
  if (l1 == 0 && l2 == 0) {
    for (i = 0; i < n; i++)
      if (!a[i]) {
        a[i] = 'a';
        while (a[i] == s[i] || a[i] == s1[i]) a[i]++;
      }
    cout << a;
  } else
    cout << -1;
  return 0;
}

### Prompt

Please create a solution in CPP 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 a[1000000];
int main() {
  int i, j, k, n, m, l1, l2;
  string s, s1;
  cin >> n >> m >> s >> s1;
  l1 = l2 = n - m;
  for (i = 0; i < n; i++) {
    if (s[i] == s1[i] && l1 && l2) {
      a[i] = s[i];
      l1--;
      l2--;
    }
  }
  for (i = 0; i < n; i++) {
    if (!a[i] && l1) {
      a[i] = s[i];
      l1--;
    }
  }
  for (i = 0; i < n; i++) {
    if (!a[i] && l2) {
      a[i] = s1[i];
      l2--;
    }
  }
  if (l1 == 0 && l2 == 0) {
    for (i = 0; i < n; i++)
      if (!a[i]) {
        a[i] = 'a';
        while (a[i] == s[i] || a[i] == s1[i]) a[i]++;
      }
    cout << a;
  } else
    cout << -1;
  return 0;
}

```