#include <bits/stdc++.h>
bool check_prime(int n) {
  for (int i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  int n, t;
  cin >> n >> t;
  string s1;
  t = n - t;
  string s2;
  cin >> s1 >> s2;
  int diff = 0;
  int same = 0;
  for (int i = 0; i < n; ++i) {
    if (s1[i] == s2[i]) {
      same++;
    } else
      diff++;
  }
  int diff1 = 0;
  int same1 = 0;
  int i;
  for (i = 0; i <= same; ++i) {
    if (diff >= 2 * (t - i)) {
      diff1 = t - i;
      same1 = i;
      break;
    }
  }
  if (i > same) {
    cout << "-1" << endl;
    return 0;
  }
  char a[n + 1];
  int count = 0;
  diff1 = 2 * diff1;
  for (int i = 0; i < n; ++i) {
    if (s1[i] == s2[i]) {
      if (same1) {
        a[i] = s1[i];
        same1--;
      } else {
        a[i] = (s1[i] == 'a') ? 'b' : 'a';
      }
    } else {
      if (diff1) {
        if (count % 2) {
          a[i] = s1[i];
          count++;
          diff1--;
        } else {
          a[i] = s2[i];
          count++;
          diff1--;
        }
      } else {
        if (s1[i] != 'a' && s2[i] != 'a') {
          a[i] = 'a';
        } else if (s1[i] != 'b' && s2[i] != 'b') {
          a[i] = 'b';
        } else if (s1[i] != 'c' && s2[i] != 'c') {
          a[i] = 'c';
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << a[i];
  }
  cout << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
bool check_prime(int n) {
  for (int i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  int n, t;
  cin >> n >> t;
  string s1;
  t = n - t;
  string s2;
  cin >> s1 >> s2;
  int diff = 0;
  int same = 0;
  for (int i = 0; i < n; ++i) {
    if (s1[i] == s2[i]) {
      same++;
    } else
      diff++;
  }
  int diff1 = 0;
  int same1 = 0;
  int i;
  for (i = 0; i <= same; ++i) {
    if (diff >= 2 * (t - i)) {
      diff1 = t - i;
      same1 = i;
      break;
    }
  }
  if (i > same) {
    cout << "-1" << endl;
    return 0;
  }
  char a[n + 1];
  int count = 0;
  diff1 = 2 * diff1;
  for (int i = 0; i < n; ++i) {
    if (s1[i] == s2[i]) {
      if (same1) {
        a[i] = s1[i];
        same1--;
      } else {
        a[i] = (s1[i] == 'a') ? 'b' : 'a';
      }
    } else {
      if (diff1) {
        if (count % 2) {
          a[i] = s1[i];
          count++;
          diff1--;
        } else {
          a[i] = s2[i];
          count++;
          diff1--;
        }
      } else {
        if (s1[i] != 'a' && s2[i] != 'a') {
          a[i] = 'a';
        } else if (s1[i] != 'b' && s2[i] != 'b') {
          a[i] = 'b';
        } else if (s1[i] != 'c' && s2[i] != 'c') {
          a[i] = 'c';
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << a[i];
  }
  cout << endl;
  return 0;
}

```