#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    long long n, k, i, p = 0;
    cin >> n >> k;
    string s1, s2, s3;
    cin >> s1 >> s2;
    for (i = 0; i < n; i++) {
      s3 += '0';
      if (s1[i] != s2[i]) p++;
    }
    if (p > 2 * k)
      cout << -1;
    else if (p <= 2 * k and p >= k) {
      long long a = (2 * k) - p, cnt = 0;
      for (i = 0; i < n; i++) {
        if (cnt == a) break;
        if (s1[i] != s2[i]) {
          if (s1[i] != 'a' and s2[i] != 'a')
            s3[i] = 'a';
          else if (s1[i] != 'b' and s2[i] != 'b')
            s3[i] = 'b';
          else if (s1[i] != 'c' and s2[i] != 'c')
            s3[i] = 'c';
          cnt++;
        }
      }
      p -= a;
      k -= a;
      cnt = 0;
      for (i = 0; i < n; i++) {
        if (cnt == (p / 2)) break;
        if (s1[i] != s2[i] and s3[i] == '0') {
          s3[i] = s1[i];
          cnt++;
        }
      }
      for (i = 0; i < n; i++) {
        if (cnt == p) break;
        if (s1[i] != s2[i] and s3[i] == '0') {
          s3[i] = s2[i];
          cnt++;
        }
      }
      for (i = 0; i < n; i++) {
        if (s3[i] == '0') s3[i] = s1[i];
      }
      cout << s3;
    } else if (k > p) {
      for (i = 0; i < n; i++) {
        if (s1[i] != s2[i]) {
          if (s1[i] != 'a' and s2[i] != 'a')
            s3[i] = 'a';
          else if (s1[i] != 'b' and s2[i] != 'b')
            s3[i] = 'b';
          else if (s1[i] != 'c' and s2[i] != 'c')
            s3[i] = 'c';
        }
      }
      long long cnt = p;
      for (i = 0; i < n; i++) {
        if (cnt == k) break;
        if (s3[i] == '0') {
          if (s1[i] != 'a' and s2[i] != 'a')
            s3[i] = 'a';
          else if (s1[i] != 'b' and s2[i] != 'b')
            s3[i] = 'b';
          else if (s1[i] != 'c' and s2[i] != 'c')
            s3[i] = 'c';
          cnt++;
        }
      }
      for (i = 0; i < n; i++) {
        if (s3[i] == '0') s3[i] = s1[i];
      }
      cout << s3;
    }
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    long long n, k, i, p = 0;
    cin >> n >> k;
    string s1, s2, s3;
    cin >> s1 >> s2;
    for (i = 0; i < n; i++) {
      s3 += '0';
      if (s1[i] != s2[i]) p++;
    }
    if (p > 2 * k)
      cout << -1;
    else if (p <= 2 * k and p >= k) {
      long long a = (2 * k) - p, cnt = 0;
      for (i = 0; i < n; i++) {
        if (cnt == a) break;
        if (s1[i] != s2[i]) {
          if (s1[i] != 'a' and s2[i] != 'a')
            s3[i] = 'a';
          else if (s1[i] != 'b' and s2[i] != 'b')
            s3[i] = 'b';
          else if (s1[i] != 'c' and s2[i] != 'c')
            s3[i] = 'c';
          cnt++;
        }
      }
      p -= a;
      k -= a;
      cnt = 0;
      for (i = 0; i < n; i++) {
        if (cnt == (p / 2)) break;
        if (s1[i] != s2[i] and s3[i] == '0') {
          s3[i] = s1[i];
          cnt++;
        }
      }
      for (i = 0; i < n; i++) {
        if (cnt == p) break;
        if (s1[i] != s2[i] and s3[i] == '0') {
          s3[i] = s2[i];
          cnt++;
        }
      }
      for (i = 0; i < n; i++) {
        if (s3[i] == '0') s3[i] = s1[i];
      }
      cout << s3;
    } else if (k > p) {
      for (i = 0; i < n; i++) {
        if (s1[i] != s2[i]) {
          if (s1[i] != 'a' and s2[i] != 'a')
            s3[i] = 'a';
          else if (s1[i] != 'b' and s2[i] != 'b')
            s3[i] = 'b';
          else if (s1[i] != 'c' and s2[i] != 'c')
            s3[i] = 'c';
        }
      }
      long long cnt = p;
      for (i = 0; i < n; i++) {
        if (cnt == k) break;
        if (s3[i] == '0') {
          if (s1[i] != 'a' and s2[i] != 'a')
            s3[i] = 'a';
          else if (s1[i] != 'b' and s2[i] != 'b')
            s3[i] = 'b';
          else if (s1[i] != 'c' and s2[i] != 'c')
            s3[i] = 'c';
          cnt++;
        }
      }
      for (i = 0; i < n; i++) {
        if (s3[i] == '0') s3[i] = s1[i];
      }
      cout << s3;
    }
  }
  return 0;
}

```