#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100100;
int n, t;
char s1[MAXN];
char s2[MAXN];
char res[MAXN];
bool used[MAXN];
char get(char c1, char c2) {
  for (char c = 'a'; c <= 'z'; c++) {
    if (c1 != c && c2 != c) {
      return c;
    }
  }
  return 'z';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> t;
  cin >> (s1 + 1);
  cin >> (s2 + 1);
  int q = 0;
  for (int i = 1; i <= n; i++) {
    if (s1[i] == s2[i]) {
      q++;
    }
  }
  int k = n - t;
  if (k <= q) {
    for (int i = 1; i <= n; i++) {
      if (s1[i] == s2[i]) {
        if (k > 0) {
          res[i] = s1[i];
          k--;
        } else {
          res[i] = get(s1[i], s2[i]);
        }
      } else {
        res[i] = get(s1[i], s2[i]);
      }
    }
  } else if (n - q >= 2 * (k - q)) {
    for (int i = 1; i <= n; i++) {
      if (s1[i] == s2[i]) {
        res[i] = s1[i];
        used[i] = true;
      }
    }
    int x = k - q;
    for (int i = 1; i <= n; i++) {
      if (used[i]) {
        continue;
      }
      if (s1[i] != s2[i]) {
        if (x > 0) {
          res[i] = s1[i];
          used[i] = true;
          x--;
        } else {
          break;
        }
      }
    }
    x = k - q;
    for (int i = 1; i <= n; i++) {
      if (used[i]) {
        continue;
      }
      if (s1[i] != s2[i]) {
        if (x > 0) {
          res[i] = s2[i];
          used[i] = true;
          x--;
        } else {
          break;
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      if (!used[i]) {
        res[i] = get(s1[i], s2[i]);
      }
    }
  } else {
    cout << -1;
    return 0;
  }
  cout << (res + 1) << "\n";
  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;
const int MAXN = 100100;
int n, t;
char s1[MAXN];
char s2[MAXN];
char res[MAXN];
bool used[MAXN];
char get(char c1, char c2) {
  for (char c = 'a'; c <= 'z'; c++) {
    if (c1 != c && c2 != c) {
      return c;
    }
  }
  return 'z';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> t;
  cin >> (s1 + 1);
  cin >> (s2 + 1);
  int q = 0;
  for (int i = 1; i <= n; i++) {
    if (s1[i] == s2[i]) {
      q++;
    }
  }
  int k = n - t;
  if (k <= q) {
    for (int i = 1; i <= n; i++) {
      if (s1[i] == s2[i]) {
        if (k > 0) {
          res[i] = s1[i];
          k--;
        } else {
          res[i] = get(s1[i], s2[i]);
        }
      } else {
        res[i] = get(s1[i], s2[i]);
      }
    }
  } else if (n - q >= 2 * (k - q)) {
    for (int i = 1; i <= n; i++) {
      if (s1[i] == s2[i]) {
        res[i] = s1[i];
        used[i] = true;
      }
    }
    int x = k - q;
    for (int i = 1; i <= n; i++) {
      if (used[i]) {
        continue;
      }
      if (s1[i] != s2[i]) {
        if (x > 0) {
          res[i] = s1[i];
          used[i] = true;
          x--;
        } else {
          break;
        }
      }
    }
    x = k - q;
    for (int i = 1; i <= n; i++) {
      if (used[i]) {
        continue;
      }
      if (s1[i] != s2[i]) {
        if (x > 0) {
          res[i] = s2[i];
          used[i] = true;
          x--;
        } else {
          break;
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      if (!used[i]) {
        res[i] = get(s1[i], s2[i]);
      }
    }
  } else {
    cout << -1;
    return 0;
  }
  cout << (res + 1) << "\n";
  return 0;
}

```