#include <bits/stdc++.h>
using namespace std;
int m[100010];
char s1[100010], s2[100010], s3[100010];
int main() {
  int n, t, f = 0, c = 0, n1, x;
  cin >> n >> t;
  scanf("%s %s", s1, s2);
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      m[i] = 1;
      c++;
    }
  }
  if (c > 2 * t) {
    cout << "-1\n";
  } else {
    if (c <= t) {
      for (int i = 0; i < n; i++) {
        if (m[i]) {
          for (int j = 0; j < 26; j++) {
            if ('a' + j != s1[i] && 'a' + j != s2[i]) {
              s3[i] = 'a' + j;
              break;
            }
          }
        } else if (t - c) {
          for (int j = 0; j < 26; j++) {
            if ('a' + j != s1[i] && 'a' + j != s2[i]) {
              s3[i] = 'a' + j;
              break;
            }
          }
          c++;
        } else {
          s3[i] = s1[i];
        }
      }
      s3[n] = '\0';
      printf("%s\n", s3);
    } else {
      for (int i = 0; i <= n; i++) {
        if (2 * i <= c && (n - t - i) >= 0 && (n - t - i) <= (n - c)) {
          n1 = i;
          x = n - t - i;
          break;
        } else if (2 * i > c || (n - t - i) < 0) {
          cout << "-1\n";
          return 0;
        }
      }
      for (int i = 0; i < n; i++) {
        if (m[i] && n1) {
          if (f)
            s3[i] = s1[i];
          else
            s3[i] = s2[i];
          n1 -= f;
          f = 1 - f;
        } else if (m[i]) {
          for (int j = 0; j < 26; j++) {
            if ('a' + j != s1[i] && 'a' + j != s2[i]) {
              s3[i] = 'a' + j;
              break;
            }
          }
        } else if (x) {
          s3[i] = s1[i];
          x--;
        } else {
          for (int j = 0; j < 26; j++) {
            if ('a' + j != s1[i] && 'a' + j != s2[i]) {
              s3[i] = 'a' + j;
              break;
            }
          }
        }
      }
      s3[n] = '\0';
      printf("%s\n", s3);
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int m[100010];
char s1[100010], s2[100010], s3[100010];
int main() {
  int n, t, f = 0, c = 0, n1, x;
  cin >> n >> t;
  scanf("%s %s", s1, s2);
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      m[i] = 1;
      c++;
    }
  }
  if (c > 2 * t) {
    cout << "-1\n";
  } else {
    if (c <= t) {
      for (int i = 0; i < n; i++) {
        if (m[i]) {
          for (int j = 0; j < 26; j++) {
            if ('a' + j != s1[i] && 'a' + j != s2[i]) {
              s3[i] = 'a' + j;
              break;
            }
          }
        } else if (t - c) {
          for (int j = 0; j < 26; j++) {
            if ('a' + j != s1[i] && 'a' + j != s2[i]) {
              s3[i] = 'a' + j;
              break;
            }
          }
          c++;
        } else {
          s3[i] = s1[i];
        }
      }
      s3[n] = '\0';
      printf("%s\n", s3);
    } else {
      for (int i = 0; i <= n; i++) {
        if (2 * i <= c && (n - t - i) >= 0 && (n - t - i) <= (n - c)) {
          n1 = i;
          x = n - t - i;
          break;
        } else if (2 * i > c || (n - t - i) < 0) {
          cout << "-1\n";
          return 0;
        }
      }
      for (int i = 0; i < n; i++) {
        if (m[i] && n1) {
          if (f)
            s3[i] = s1[i];
          else
            s3[i] = s2[i];
          n1 -= f;
          f = 1 - f;
        } else if (m[i]) {
          for (int j = 0; j < 26; j++) {
            if ('a' + j != s1[i] && 'a' + j != s2[i]) {
              s3[i] = 'a' + j;
              break;
            }
          }
        } else if (x) {
          s3[i] = s1[i];
          x--;
        } else {
          for (int j = 0; j < 26; j++) {
            if ('a' + j != s1[i] && 'a' + j != s2[i]) {
              s3[i] = 'a' + j;
              break;
            }
          }
        }
      }
      s3[n] = '\0';
      printf("%s\n", s3);
    }
  }
  return 0;
}

```