#include <bits/stdc++.h>
using namespace std;
template <typename T, typename TT>
ostream& operator<<(ostream& os, const pair<T, TT>& t) {
  return os << t.first << " " << t.second;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& t) {
  for (auto& i : t) os << i << " ";
  return os;
}
char ne(char a, char b) {
  if (a != 'a' && b != 'a') return 'a';
  if (a != 'b' && b != 'b')
    return 'b';
  else
    return 'c';
}
int main() {
  long long n, t, d = 0;
  scanf("%lld %lld", &n, &t);
  string s1;
  string s2;
  cin >> s1 >> s2;
  for (long long i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      d++;
    }
  }
  if (d > 2 * t) {
    cout << "-1";
    return 0;
  }
  if (d == t) {
    for (long long i = 0; i < n; i++) {
      if (s1[i] != s2[i]) {
        t--;
        s1[i] = ne(s1[i], s2[i]);
      }
    }
  } else if (d < t) {
    long long k = t - d;
    for (long long i = 0; i < n; i++) {
      if (s1[i] != s2[i] && d != 0) {
        d--;
        s1[i] = ne(s1[i], s2[i]);
      }
      if (s1[i] == s2[i] && k != 0) {
        k--;
        s1[i] = ne(s1[i], s2[i]);
      }
    }
  } else {
    long long k = d - t;
    long long l = t - k;
    for (long long i = 0; i < n; i++) {
      if (s1[i] != s2[i] && k != 0) {
        k--;
        s1[i] = s2[i];
      } else if (s1[i] != s2[i] && l != 0) {
        l--;
        s1[i] = ne(s1[i], s2[i]);
      }
    }
  }
  cout << s1;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
template <typename T, typename TT>
ostream& operator<<(ostream& os, const pair<T, TT>& t) {
  return os << t.first << " " << t.second;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& t) {
  for (auto& i : t) os << i << " ";
  return os;
}
char ne(char a, char b) {
  if (a != 'a' && b != 'a') return 'a';
  if (a != 'b' && b != 'b')
    return 'b';
  else
    return 'c';
}
int main() {
  long long n, t, d = 0;
  scanf("%lld %lld", &n, &t);
  string s1;
  string s2;
  cin >> s1 >> s2;
  for (long long i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      d++;
    }
  }
  if (d > 2 * t) {
    cout << "-1";
    return 0;
  }
  if (d == t) {
    for (long long i = 0; i < n; i++) {
      if (s1[i] != s2[i]) {
        t--;
        s1[i] = ne(s1[i], s2[i]);
      }
    }
  } else if (d < t) {
    long long k = t - d;
    for (long long i = 0; i < n; i++) {
      if (s1[i] != s2[i] && d != 0) {
        d--;
        s1[i] = ne(s1[i], s2[i]);
      }
      if (s1[i] == s2[i] && k != 0) {
        k--;
        s1[i] = ne(s1[i], s2[i]);
      }
    }
  } else {
    long long k = d - t;
    long long l = t - k;
    for (long long i = 0; i < n; i++) {
      if (s1[i] != s2[i] && k != 0) {
        k--;
        s1[i] = s2[i];
      } else if (s1[i] != s2[i] && l != 0) {
        l--;
        s1[i] = ne(s1[i], s2[i]);
      }
    }
  }
  cout << s1;
  return 0;
}

```