#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f;
const int INF = 0x3f3f3f3f;
int n, t;
string a, b;
int main() {
  while (cin >> n >> t) {
    cin >> a >> b;
    int same = 0, diff = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] == b[i])
        same++;
      else
        diff++;
    }
    if (same + diff == t) {
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < 26; j++) {
          char c = (char)(j + 97);
          if (c != a[i] && c != b[i]) {
            cout << c;
            break;
          }
        }
      }
    } else {
      if (diff <= t) {
        int remain = t - diff;
        char c;
        for (int i = 0; i < n; i++) {
          for (int j = 0; j < 26; j++) {
            c = (char)(j + 97);
            if (c != a[i] && c != b[i]) break;
          }
          if (a[i] != b[i])
            cout << c;
          else {
            if (remain) {
              remain--;
              cout << c;
            } else
              cout << a[i];
          }
        }
      } else {
        if ((diff + 1) / 2 > t)
          cout << -1;
        else {
          int flag = 0;
          int remain = 2 * diff - 2 * t;
          for (int i = 0; i < n; i++) {
            if (a[i] != b[i]) {
              if (remain) {
                if (flag)
                  cout << a[i];
                else
                  cout << b[i];
                flag = !flag;
                remain--;
              } else {
                for (int j = 0; j < 26; j++) {
                  char c = (char)(j + 97);
                  if (c != a[i] && c != b[i]) {
                    cout << c;
                    break;
                  }
                }
              }
            } else
              cout << a[i];
          }
        }
      }
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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 int inf = 0x3f;
const int INF = 0x3f3f3f3f;
int n, t;
string a, b;
int main() {
  while (cin >> n >> t) {
    cin >> a >> b;
    int same = 0, diff = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] == b[i])
        same++;
      else
        diff++;
    }
    if (same + diff == t) {
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < 26; j++) {
          char c = (char)(j + 97);
          if (c != a[i] && c != b[i]) {
            cout << c;
            break;
          }
        }
      }
    } else {
      if (diff <= t) {
        int remain = t - diff;
        char c;
        for (int i = 0; i < n; i++) {
          for (int j = 0; j < 26; j++) {
            c = (char)(j + 97);
            if (c != a[i] && c != b[i]) break;
          }
          if (a[i] != b[i])
            cout << c;
          else {
            if (remain) {
              remain--;
              cout << c;
            } else
              cout << a[i];
          }
        }
      } else {
        if ((diff + 1) / 2 > t)
          cout << -1;
        else {
          int flag = 0;
          int remain = 2 * diff - 2 * t;
          for (int i = 0; i < n; i++) {
            if (a[i] != b[i]) {
              if (remain) {
                if (flag)
                  cout << a[i];
                else
                  cout << b[i];
                flag = !flag;
                remain--;
              } else {
                for (int j = 0; j < 26; j++) {
                  char c = (char)(j + 97);
                  if (c != a[i] && c != b[i]) {
                    cout << c;
                    break;
                  }
                }
              }
            } else
              cout << a[i];
          }
        }
      }
    }
    cout << endl;
  }
  return 0;
}

```