#include <bits/stdc++.h>
using namespace std;
char other(char a, char b) {
  char arr[] = {'a', 'b', 'c'};
  for (int i = 0; i < 3; i++) {
    if (arr[i] != a && arr[i] != b) {
      return arr[i];
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, t;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  int c, d;
  vector<int> posa, posb;
  char ans[n];
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      posa.push_back(i);
    } else {
      posb.push_back(i);
    }
  }
  if (t < (posb.size() + 2 - 1) / 2) {
    cout << -1;
  } else {
    if (posb.size() % 2 == 1) {
      int p = posb[0];
      ans[p] = other(a[p], b[p]);
      t--;
    }
    int i = posb.size() % 2;
    while (i < posb.size()) {
      int p = posb[i];
      if (i % 2 == 0) {
        ans[p] = a[p];
        t--;
      } else {
        ans[p] = b[p];
      }
      i++;
    }
    i = posb.size() % 2;
    while (i < posb.size() && t > 0) {
      int p = posb[i];
      int k = posb[i + 1];
      ans[p] = other(a[p], b[p]);
      ans[k] = other(a[k], b[k]);
      t--;
      i += 2;
    }
    i = 0;
    while (i < posa.size() && t > 0) {
      int p = posa[i];
      ans[p] = other(a[p], b[p]);
      t--;
      i++;
    }
    while (i < posa.size()) {
      int p = posa[i];
      ans[p] = a[p];
      i++;
    }
    for (int i = 0; i < n; i++) {
      cout << ans[i];
    }
    cout << '\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;
char other(char a, char b) {
  char arr[] = {'a', 'b', 'c'};
  for (int i = 0; i < 3; i++) {
    if (arr[i] != a && arr[i] != b) {
      return arr[i];
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, t;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  int c, d;
  vector<int> posa, posb;
  char ans[n];
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      posa.push_back(i);
    } else {
      posb.push_back(i);
    }
  }
  if (t < (posb.size() + 2 - 1) / 2) {
    cout << -1;
  } else {
    if (posb.size() % 2 == 1) {
      int p = posb[0];
      ans[p] = other(a[p], b[p]);
      t--;
    }
    int i = posb.size() % 2;
    while (i < posb.size()) {
      int p = posb[i];
      if (i % 2 == 0) {
        ans[p] = a[p];
        t--;
      } else {
        ans[p] = b[p];
      }
      i++;
    }
    i = posb.size() % 2;
    while (i < posb.size() && t > 0) {
      int p = posb[i];
      int k = posb[i + 1];
      ans[p] = other(a[p], b[p]);
      ans[k] = other(a[k], b[k]);
      t--;
      i += 2;
    }
    i = 0;
    while (i < posa.size() && t > 0) {
      int p = posa[i];
      ans[p] = other(a[p], b[p]);
      t--;
      i++;
    }
    while (i < posa.size()) {
      int p = posa[i];
      ans[p] = a[p];
      i++;
    }
    for (int i = 0; i < n; i++) {
      cout << ans[i];
    }
    cout << '\n';
  }
  return 0;
}

```