#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k, p1 = 0, p2 = 0;
  string s, p, t = "";
  cin >> n >> k;
  cin >> s >> p;
  p1 = n - k;
  p2 = n - k;
  for (int i = 0; i < s.length(); i++) t += '-';
  for (int i = 0; i < n; i++) {
    if (s[i] == p[i] && p1 > 0 && p2 > 0) {
      p1--;
      p2--;
      t[i] = s[i];
    }
  }
  for (int i = 0; i < p.length(); i++) {
    if (t[i] == '-') {
      if (p1 > 0) {
        t[i] = s[i];
        p1--;
      } else if (p2 > 0) {
        t[i] = p[i];
        p2--;
      } else {
        for (int l = 0; l < 26; l++) {
          if (char(l + 'a') != p[i] && s[i] != char(l + 'a')) {
            t[i] = char(l + 'a');
            break;
          }
        }
      }
    }
  }
  if (p1 == 0 && p2 == 0)
    cout << t;
  else
    cout << -1;
}

### 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;
int main() {
  int n, k, p1 = 0, p2 = 0;
  string s, p, t = "";
  cin >> n >> k;
  cin >> s >> p;
  p1 = n - k;
  p2 = n - k;
  for (int i = 0; i < s.length(); i++) t += '-';
  for (int i = 0; i < n; i++) {
    if (s[i] == p[i] && p1 > 0 && p2 > 0) {
      p1--;
      p2--;
      t[i] = s[i];
    }
  }
  for (int i = 0; i < p.length(); i++) {
    if (t[i] == '-') {
      if (p1 > 0) {
        t[i] = s[i];
        p1--;
      } else if (p2 > 0) {
        t[i] = p[i];
        p2--;
      } else {
        for (int l = 0; l < 26; l++) {
          if (char(l + 'a') != p[i] && s[i] != char(l + 'a')) {
            t[i] = char(l + 'a');
            break;
          }
        }
      }
    }
  }
  if (p1 == 0 && p2 == 0)
    cout << t;
  else
    cout << -1;
}

```