#include <bits/stdc++.h>
using namespace std;
char getDiffChar(char a, char b) {
  if (a != 'a' && b != 'a') {
    return 'a';
  }
  if (a != 'b' && b != 'b') {
    return 'b';
  }
  return 'c';
}
int main() {
  std::ios_base::sync_with_stdio(false);
  int n, t;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  int dif = 0;
  for (int i = 0; i < int(n); i++) {
    dif += int(a[i] != b[i]);
  }
  string ans = string(n, '*');
  if (t >= dif) {
    for (int i = 0; i < int(n); i++) {
      if (t == 0) {
        break;
      }
      if (a[i] != b[i]) {
        ans[i] = getDiffChar(a[i], b[i]);
        t--;
      }
    }
    for (int i = 0; i < int(n); i++) {
      if (t == 0) {
        break;
      }
      if (a[i] == b[i]) {
        ans[i] = getDiffChar(a[i], b[i]);
        t--;
      }
    }
    for (int i = 0; i < int(n); i++) {
      if (ans[i] == '*') {
        ans[i] = a[i];
      }
    }
  } else if (t >= dif / 2 + dif % 2) {
    int r = 2 * t - dif;
    for (int i = 0; i < int(n); i++) {
      if (r == 0) {
        break;
      }
      if (a[i] != b[i]) {
        ans[i] = getDiffChar(a[i], b[i]);
        r--;
      }
    }
    int k = t - (2 * t - dif);
    for (int i = 0; i < int(n); i++) {
      if (k == 0) {
        break;
      }
      if (ans[i] == '*' && a[i] != b[i]) {
        ans[i] = b[i];
        k--;
      }
    }
    k = t - (2 * t - dif);
    for (int i = 0; i < int(n); i++) {
      if (k == 0) {
        break;
      }
      if (ans[i] == '*' && a[i] != b[i]) {
        ans[i] = a[i];
        k--;
      }
    }
    for (int i = 0; i < int(n); i++) {
      if (ans[i] == '*') {
        ans[i] = a[i];
      }
    }
  } else {
    ans = "-1";
  }
  cout << ans;
  return 0;
}

### Prompt

Generate 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;
char getDiffChar(char a, char b) {
  if (a != 'a' && b != 'a') {
    return 'a';
  }
  if (a != 'b' && b != 'b') {
    return 'b';
  }
  return 'c';
}
int main() {
  std::ios_base::sync_with_stdio(false);
  int n, t;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  int dif = 0;
  for (int i = 0; i < int(n); i++) {
    dif += int(a[i] != b[i]);
  }
  string ans = string(n, '*');
  if (t >= dif) {
    for (int i = 0; i < int(n); i++) {
      if (t == 0) {
        break;
      }
      if (a[i] != b[i]) {
        ans[i] = getDiffChar(a[i], b[i]);
        t--;
      }
    }
    for (int i = 0; i < int(n); i++) {
      if (t == 0) {
        break;
      }
      if (a[i] == b[i]) {
        ans[i] = getDiffChar(a[i], b[i]);
        t--;
      }
    }
    for (int i = 0; i < int(n); i++) {
      if (ans[i] == '*') {
        ans[i] = a[i];
      }
    }
  } else if (t >= dif / 2 + dif % 2) {
    int r = 2 * t - dif;
    for (int i = 0; i < int(n); i++) {
      if (r == 0) {
        break;
      }
      if (a[i] != b[i]) {
        ans[i] = getDiffChar(a[i], b[i]);
        r--;
      }
    }
    int k = t - (2 * t - dif);
    for (int i = 0; i < int(n); i++) {
      if (k == 0) {
        break;
      }
      if (ans[i] == '*' && a[i] != b[i]) {
        ans[i] = b[i];
        k--;
      }
    }
    k = t - (2 * t - dif);
    for (int i = 0; i < int(n); i++) {
      if (k == 0) {
        break;
      }
      if (ans[i] == '*' && a[i] != b[i]) {
        ans[i] = a[i];
        k--;
      }
    }
    for (int i = 0; i < int(n); i++) {
      if (ans[i] == '*') {
        ans[i] = a[i];
      }
    }
  } else {
    ans = "-1";
  }
  cout << ans;
  return 0;
}

```