#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 4;
string a, b;
char s[N];
int n, t;
int main() {
  cin >> n >> t;
  cin >> a >> b;
  int x = n - t, c = 0;
  for (int i = 0; i < n; i++)
    if (a[i] == b[i]) c++;
  if (x <= c) {
    for (int i = 0; i < n; i++) {
      if (a[i] == b[i] && x) {
        cout << a[i];
        x--;
        continue;
      } else
        for (int j = 0; j < 26; j++)
          if (char(j + 'a') != a[i] && char(j + 'a') != b[i]) {
            cout << char(j + 'a');
            break;
          }
    }
    return 0;
  }
  if (x > c) {
    x -= c;
    if (x * 2 > (n - c)) return puts("-1"), 0;
    int c1 = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        s[i] = a[i];
        a[i] = '!';
        c1++;
      }
      if (c1 == x) break;
    }
    c1 = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] != b[i] && a[i] != '!') {
        s[i] = b[i];
        b[i] = '!';
        c1++;
      }
      if (c1 == x) break;
    }
    for (int i = 0; i < n; i++) {
      if (a[i] != '!' && b[i] != '!') {
        if (a[i] == b[i]) {
          s[i] = a[i];
          x--;
        } else {
          for (int j = 0; j < 26; j++)
            if (char(j + 'a') != a[i] && char(j + 'a') != b[i]) {
              s[i] = char(j + 'a');
              break;
            }
        }
      }
    }
    for (int i = 0; i < n; i++) cout << s[i];
  }
  return 0;
}

### 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;
const int N = 1e6 + 4;
string a, b;
char s[N];
int n, t;
int main() {
  cin >> n >> t;
  cin >> a >> b;
  int x = n - t, c = 0;
  for (int i = 0; i < n; i++)
    if (a[i] == b[i]) c++;
  if (x <= c) {
    for (int i = 0; i < n; i++) {
      if (a[i] == b[i] && x) {
        cout << a[i];
        x--;
        continue;
      } else
        for (int j = 0; j < 26; j++)
          if (char(j + 'a') != a[i] && char(j + 'a') != b[i]) {
            cout << char(j + 'a');
            break;
          }
    }
    return 0;
  }
  if (x > c) {
    x -= c;
    if (x * 2 > (n - c)) return puts("-1"), 0;
    int c1 = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        s[i] = a[i];
        a[i] = '!';
        c1++;
      }
      if (c1 == x) break;
    }
    c1 = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] != b[i] && a[i] != '!') {
        s[i] = b[i];
        b[i] = '!';
        c1++;
      }
      if (c1 == x) break;
    }
    for (int i = 0; i < n; i++) {
      if (a[i] != '!' && b[i] != '!') {
        if (a[i] == b[i]) {
          s[i] = a[i];
          x--;
        } else {
          for (int j = 0; j < 26; j++)
            if (char(j + 'a') != a[i] && char(j + 'a') != b[i]) {
              s[i] = char(j + 'a');
              break;
            }
        }
      }
    }
    for (int i = 0; i < n; i++) cout << s[i];
  }
  return 0;
}

```