#include <bits/stdc++.h>
using namespace std;
char diferit(vector<char> v) {
  vector<int> fv(26, 0);
  for (char ch : v) fv[ch - 'a'] = 1;
  for (int i = 25; i >= 0; i--)
    if (fv[i] == 0) return 'a' + i;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s, t, sol;
  int n, k;
  cin >> n >> k >> s >> t;
  k = n - k;
  int i, j, z;
  for (i = 0; i < n; i++) sol += diferit({s[i], t[i]});
  int dif = 0;
  for (i = 0; i < n; i++)
    if (s[i] != t[i]) dif++;
  dif /= 2;
  dif = min(dif, k);
  int a = 0, b = 0;
  for (i = 0; i < n; i++)
    if (s[i] != t[i]) {
      if (a < dif) {
        sol[i] = s[i];
        a++;
      } else if (b < dif) {
        sol[i] = t[i];
        b++;
      }
    }
  for (i = 0; i < n; i++)
    if (s[i] == t[i] && a < k) {
      sol[i] = s[i];
      a++;
      b++;
    }
  int da = 0, db = 0;
  for (i = 0; i < n; i++) {
    if (sol[i] == s[i]) da++;
    if (sol[i] == t[i]) db++;
  }
  if (da == k && db == k)
    cout << sol;
  else
    cout << -1;
  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;
char diferit(vector<char> v) {
  vector<int> fv(26, 0);
  for (char ch : v) fv[ch - 'a'] = 1;
  for (int i = 25; i >= 0; i--)
    if (fv[i] == 0) return 'a' + i;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s, t, sol;
  int n, k;
  cin >> n >> k >> s >> t;
  k = n - k;
  int i, j, z;
  for (i = 0; i < n; i++) sol += diferit({s[i], t[i]});
  int dif = 0;
  for (i = 0; i < n; i++)
    if (s[i] != t[i]) dif++;
  dif /= 2;
  dif = min(dif, k);
  int a = 0, b = 0;
  for (i = 0; i < n; i++)
    if (s[i] != t[i]) {
      if (a < dif) {
        sol[i] = s[i];
        a++;
      } else if (b < dif) {
        sol[i] = t[i];
        b++;
      }
    }
  for (i = 0; i < n; i++)
    if (s[i] == t[i] && a < k) {
      sol[i] = s[i];
      a++;
      b++;
    }
  int da = 0, db = 0;
  for (i = 0; i < n; i++) {
    if (sol[i] == s[i]) da++;
    if (sol[i] == t[i]) db++;
  }
  if (da == k && db == k)
    cout << sol;
  else
    cout << -1;
  return 0;
}

```