#include <bits/stdc++.h>
using namespace std;
int n, t, eq, dif, k, swp;
string s1, s2, s3;
int f(string a, string b) {
  int dif = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] != b[i]) dif++;
  }
  return dif;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> t;
  dif = 0;
  cin >> s1;
  cin >> s2;
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      dif++;
    }
  }
  k = 0;
  s3 = "";
  swp = 0;
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i] || k == n - t) {
      if (s1[i] != 'a' && s2[i] != 'a') {
        s3 += 'a';
      } else if (s1[i] == 'a' && s2[i] == 'b') {
        s3 += 'c';
      } else if (s1[i] == 'b' && s2[i] == 'a') {
        s3 += 'c';
      } else if (s1[i] == 'a') {
        s3 += 'b';
      } else if (s2[i] == 'a') {
        s3 += 'b';
      } else {
        s3 += 'a';
      }
    } else {
      s3 += s1[i];
      k++;
    }
  }
  int l1 = f(s1, s3);
  int zk = 0;
  for (int i = 0; i < l1 - t; i++) {
    for (int j = zk; j < n; j++) {
      if (s3[j] != s1[j] && s3[j] != s2[j]) {
        s3[j] = s1[j];
        break;
      }
      zk++;
    }
  }
  int l2 = f(s2, s3);
  zk = 0;
  for (int i = 0; i < l2 - t; i++) {
    for (int j = zk; j < n; j++) {
      if (s3[j] != s1[j] && s3[j] != s2[j]) {
        s3[j] = s2[j];
        break;
      }
      zk++;
    }
  }
  l1 = f(s1, s3);
  l2 = f(s2, s3);
  if (l1 == l2 && l2 == t) {
    cout << s3 << "\n";
  } else {
    cout << "-1\n";
  }
  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;
int n, t, eq, dif, k, swp;
string s1, s2, s3;
int f(string a, string b) {
  int dif = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] != b[i]) dif++;
  }
  return dif;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> t;
  dif = 0;
  cin >> s1;
  cin >> s2;
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      dif++;
    }
  }
  k = 0;
  s3 = "";
  swp = 0;
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i] || k == n - t) {
      if (s1[i] != 'a' && s2[i] != 'a') {
        s3 += 'a';
      } else if (s1[i] == 'a' && s2[i] == 'b') {
        s3 += 'c';
      } else if (s1[i] == 'b' && s2[i] == 'a') {
        s3 += 'c';
      } else if (s1[i] == 'a') {
        s3 += 'b';
      } else if (s2[i] == 'a') {
        s3 += 'b';
      } else {
        s3 += 'a';
      }
    } else {
      s3 += s1[i];
      k++;
    }
  }
  int l1 = f(s1, s3);
  int zk = 0;
  for (int i = 0; i < l1 - t; i++) {
    for (int j = zk; j < n; j++) {
      if (s3[j] != s1[j] && s3[j] != s2[j]) {
        s3[j] = s1[j];
        break;
      }
      zk++;
    }
  }
  int l2 = f(s2, s3);
  zk = 0;
  for (int i = 0; i < l2 - t; i++) {
    for (int j = zk; j < n; j++) {
      if (s3[j] != s1[j] && s3[j] != s2[j]) {
        s3[j] = s2[j];
        break;
      }
      zk++;
    }
  }
  l1 = f(s1, s3);
  l2 = f(s2, s3);
  if (l1 == l2 && l2 == t) {
    cout << s3 << "\n";
  } else {
    cout << "-1\n";
  }
  return 0;
}

```