#include <bits/stdc++.h>
using namespace std;
int n, k, m;
signed main() {
  cin >> n >> m;
  m = n - m;
  string a, b;
  cin >> a >> b;
  vector<int> rez(n, 0);
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      if (m) {
        rez[i] = 3;
        m--;
      } else {
        rez[i] = 4;
      }
    }
  }
  int m1 = m, m2 = m;
  for (int i = 0; i < n; i++) {
    if (rez[i] != 0) continue;
    if (m1) {
      rez[i] = 1;
      m1--;
    } else if (m2) {
      rez[i] = 2;
      m2--;
    }
  }
  if (m1 || m2) {
    cout << -1;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (rez[i] & 1) {
      cout << a[i];
    } else if (rez[i] % 4 == 0) {
      char x = 'a';
      while (x == a[i] || x == b[i]) x++;
      cout << x;
    } else if (rez[i] == 2) {
      cout << b[i];
    }
  }
}

### Prompt

Please provide a Cpp coded solution 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;
int n, k, m;
signed main() {
  cin >> n >> m;
  m = n - m;
  string a, b;
  cin >> a >> b;
  vector<int> rez(n, 0);
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      if (m) {
        rez[i] = 3;
        m--;
      } else {
        rez[i] = 4;
      }
    }
  }
  int m1 = m, m2 = m;
  for (int i = 0; i < n; i++) {
    if (rez[i] != 0) continue;
    if (m1) {
      rez[i] = 1;
      m1--;
    } else if (m2) {
      rez[i] = 2;
      m2--;
    }
  }
  if (m1 || m2) {
    cout << -1;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (rez[i] & 1) {
      cout << a[i];
    } else if (rez[i] % 4 == 0) {
      char x = 'a';
      while (x == a[i] || x == b[i]) x++;
      cout << x;
    } else if (rez[i] == 2) {
      cout << b[i];
    }
  }
}

```