#include <bits/stdc++.h>
using namespace std;
const unsigned long long MOD = 1e9 + 7;
string s1, s2, s3;
long long n, t, p1 = 0, p2 = 0;
char foo(char ch1, char ch2) {
  if (('a' != ch1) && ('a' != ch2)) return 'a';
  if (('b' != ch1) && ('b' != ch2)) return 'b';
  if (('c' != ch1) && ('c' != ch2)) return 'c';
}
int main() {
  cin >> n >> t >> s1 >> s2;
  s3 = s1;
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      p1++;
      p2++;
    } else {
      if (p1 < p2) {
        p1++;
        s3[i] = s1[i];
      } else {
        p2++;
        s3[i] = s2[i];
      }
    }
  }
  if ((t < n - p1) || (t < n - p2)) {
    cout << -1;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      continue;
    } else {
      if ((s2[i] == s3[i]) && (n - p2 < t)) {
        s3[i] = foo(s1[i], s2[i]);
        p2--;
      } else {
        if ((s1[i] == s3[i]) && (n - p1 < t)) {
          s3[i] = foo(s1[i], s2[i]);
          p1--;
        }
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if ((s2[i] == s3[i]) && (s1[i] == s3[i]) &&
        ((n - p1 < t) || (n - p2 < t))) {
      s3[i] = foo(s1[i], s2[i]);
      p2--;
      p1--;
    }
  }
  for (int i = 0; i < n; i++) {
    if ((n - p1 > t) && (s1[i] != s2[i]) && (s3[i] != s1[i]) &&
        (s3[i] != s2[i])) {
      s3[i] = s1[i];
      p1++;
    }
    if ((n - p2 > t) && (s1[i] != s2[i]) && (s3[i] != s1[i]) &&
        (s3[i] != s2[i])) {
      s3[i] = s2[i];
      p2++;
    }
  }
  if ((n - p1 != t) || (n - p2 != t)) {
    cout << -1;
    return 0;
  }
  cout << s3;
  return 0;
}

### Prompt

Please formulate 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;
const unsigned long long MOD = 1e9 + 7;
string s1, s2, s3;
long long n, t, p1 = 0, p2 = 0;
char foo(char ch1, char ch2) {
  if (('a' != ch1) && ('a' != ch2)) return 'a';
  if (('b' != ch1) && ('b' != ch2)) return 'b';
  if (('c' != ch1) && ('c' != ch2)) return 'c';
}
int main() {
  cin >> n >> t >> s1 >> s2;
  s3 = s1;
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      p1++;
      p2++;
    } else {
      if (p1 < p2) {
        p1++;
        s3[i] = s1[i];
      } else {
        p2++;
        s3[i] = s2[i];
      }
    }
  }
  if ((t < n - p1) || (t < n - p2)) {
    cout << -1;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      continue;
    } else {
      if ((s2[i] == s3[i]) && (n - p2 < t)) {
        s3[i] = foo(s1[i], s2[i]);
        p2--;
      } else {
        if ((s1[i] == s3[i]) && (n - p1 < t)) {
          s3[i] = foo(s1[i], s2[i]);
          p1--;
        }
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if ((s2[i] == s3[i]) && (s1[i] == s3[i]) &&
        ((n - p1 < t) || (n - p2 < t))) {
      s3[i] = foo(s1[i], s2[i]);
      p2--;
      p1--;
    }
  }
  for (int i = 0; i < n; i++) {
    if ((n - p1 > t) && (s1[i] != s2[i]) && (s3[i] != s1[i]) &&
        (s3[i] != s2[i])) {
      s3[i] = s1[i];
      p1++;
    }
    if ((n - p2 > t) && (s1[i] != s2[i]) && (s3[i] != s1[i]) &&
        (s3[i] != s2[i])) {
      s3[i] = s2[i];
      p2++;
    }
  }
  if ((n - p1 != t) || (n - p2 != t)) {
    cout << -1;
    return 0;
  }
  cout << s3;
  return 0;
}

```