#include <bits/stdc++.h>
using namespace std;
int main() {
  int diff = 0, length, t1, index, index2;
  cin >> length >> t1;
  char p[100100], s[100100], t[100100];
  cin >> s >> t;
  for (int i = 0; i < length; i++) {
    if (s[i] != t[i]) {
      diff++;
    }
  }
  if (t1 < (diff + 1) / 2)
    cout << -1 << endl;
  else {
    if (t1 <= diff) {
      bool flag = false;
      index = (diff - t1) * 2;
      for (int i = 0; i < length; i++) {
        if (s[i] != t[i]) {
          if (index > 0) {
            if (flag) {
              flag = false;
              p[i] = s[i];
            } else {
              flag = true;
              p[i] = t[i];
            }
            index--;
          } else {
            if ('a' != s[i] && 'a' != t[i])
              p[i] = 'a';
            else if ('b' != s[i] && 'b' != t[i])
              p[i] = 'b';
            else
              p[i] = 'c';
          }
        } else {
          p[i] = s[i];
        }
      }
    } else {
      bool flag = false;
      index2 = t1 - diff;
      for (int i = 0; i < length; i++) {
        if (s[i] != t[i]) {
          if ('a' != s[i] && 'a' != t[i])
            p[i] = 'a';
          else if ('b' != s[i] && 'b' != t[i])
            p[i] = 'b';
          else
            p[i] = 'c';
        } else {
          if (index2 > 0) {
            if ('a' != s[i])
              p[i] = 'a';
            else
              p[i] = 'b';
            index2--;
          } else
            p[i] = s[i];
        }
      }
    }
    for (int i = 0; i < length; i++) cout << p[i];
    cout << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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 diff = 0, length, t1, index, index2;
  cin >> length >> t1;
  char p[100100], s[100100], t[100100];
  cin >> s >> t;
  for (int i = 0; i < length; i++) {
    if (s[i] != t[i]) {
      diff++;
    }
  }
  if (t1 < (diff + 1) / 2)
    cout << -1 << endl;
  else {
    if (t1 <= diff) {
      bool flag = false;
      index = (diff - t1) * 2;
      for (int i = 0; i < length; i++) {
        if (s[i] != t[i]) {
          if (index > 0) {
            if (flag) {
              flag = false;
              p[i] = s[i];
            } else {
              flag = true;
              p[i] = t[i];
            }
            index--;
          } else {
            if ('a' != s[i] && 'a' != t[i])
              p[i] = 'a';
            else if ('b' != s[i] && 'b' != t[i])
              p[i] = 'b';
            else
              p[i] = 'c';
          }
        } else {
          p[i] = s[i];
        }
      }
    } else {
      bool flag = false;
      index2 = t1 - diff;
      for (int i = 0; i < length; i++) {
        if (s[i] != t[i]) {
          if ('a' != s[i] && 'a' != t[i])
            p[i] = 'a';
          else if ('b' != s[i] && 'b' != t[i])
            p[i] = 'b';
          else
            p[i] = 'c';
        } else {
          if (index2 > 0) {
            if ('a' != s[i])
              p[i] = 'a';
            else
              p[i] = 'b';
            index2--;
          } else
            p[i] = s[i];
        }
      }
    }
    for (int i = 0; i < length; i++) cout << p[i];
    cout << endl;
  }
  return 0;
}

```