#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n, t, t1, t2, diferentes = 0, iguales = 0, i, j;
  string cad1, cad2, ans = "";
  cin >> n >> t;
  cin >> cad1 >> cad2;
  vector<int> dif(n, 0);
  for (i = 0; i < n; i++) {
    if (cad1[i] == cad2[i])
      iguales++;
    else {
      diferentes++;
      dif[i] = 1;
    }
  }
  if (diferentes > 2 * t)
    cout << -1 << '\n';
  else {
    ans = "";
    for (i = 0; i < n; i++) {
      ans.push_back(cad1[i]);
    }
    if (diferentes <= t) {
      for (i = 0; i < n; i++) {
        if (dif[i] == 1) {
          t--;
          for (j = 0; j < 3; j++) {
            if (cad1[i] != 'a' + j && cad2[i] != 'a' + j) {
              ans[i] = 'a' + j;
              break;
            }
          }
        }
      }
      for (i = 0; i < n; i++) {
        if (dif[i] == 0 && t > 0) {
          t--;
          for (j = 0; j < 2; j++) {
            if (cad1[i] != 'a' + j && cad2[i] != 'a' + j) {
              ans[i] = 'a' + j;
              break;
            }
          }
        }
      }
    } else {
      t1 = diferentes - t;
      t2 = t1;
      t = t - t1;
      for (i = 0; i < n; i++) {
        if (dif[i] == 1 && t1 > 0) {
          dif[i] = 2;
          t1--;
          ans[i] = cad2[i];
        }
      }
      for (i = 0; i < n; i++) {
        if (dif[i] == 1 && t2 > 0) {
          dif[i] = 2;
          t2--;
          ans[i] = cad1[i];
        }
      }
      for (i = 0; i < n; i++) {
        if (dif[i] == 1 && t > 0) {
          dif[i] = 2;
          t--;
          for (j = 0; j < 3; j++) {
            if (cad1[i] != 'a' + j && cad2[i] != 'a' + j) {
              ans[i] = 'a' + j;
              break;
            }
          }
        }
      }
    }
    cout << ans << '\n';
  }
}

### 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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n, t, t1, t2, diferentes = 0, iguales = 0, i, j;
  string cad1, cad2, ans = "";
  cin >> n >> t;
  cin >> cad1 >> cad2;
  vector<int> dif(n, 0);
  for (i = 0; i < n; i++) {
    if (cad1[i] == cad2[i])
      iguales++;
    else {
      diferentes++;
      dif[i] = 1;
    }
  }
  if (diferentes > 2 * t)
    cout << -1 << '\n';
  else {
    ans = "";
    for (i = 0; i < n; i++) {
      ans.push_back(cad1[i]);
    }
    if (diferentes <= t) {
      for (i = 0; i < n; i++) {
        if (dif[i] == 1) {
          t--;
          for (j = 0; j < 3; j++) {
            if (cad1[i] != 'a' + j && cad2[i] != 'a' + j) {
              ans[i] = 'a' + j;
              break;
            }
          }
        }
      }
      for (i = 0; i < n; i++) {
        if (dif[i] == 0 && t > 0) {
          t--;
          for (j = 0; j < 2; j++) {
            if (cad1[i] != 'a' + j && cad2[i] != 'a' + j) {
              ans[i] = 'a' + j;
              break;
            }
          }
        }
      }
    } else {
      t1 = diferentes - t;
      t2 = t1;
      t = t - t1;
      for (i = 0; i < n; i++) {
        if (dif[i] == 1 && t1 > 0) {
          dif[i] = 2;
          t1--;
          ans[i] = cad2[i];
        }
      }
      for (i = 0; i < n; i++) {
        if (dif[i] == 1 && t2 > 0) {
          dif[i] = 2;
          t2--;
          ans[i] = cad1[i];
        }
      }
      for (i = 0; i < n; i++) {
        if (dif[i] == 1 && t > 0) {
          dif[i] = 2;
          t--;
          for (j = 0; j < 3; j++) {
            if (cad1[i] != 'a' + j && cad2[i] != 'a' + j) {
              ans[i] = 'a' + j;
              break;
            }
          }
        }
      }
    }
    cout << ans << '\n';
  }
}

```