#include <bits/stdc++.h>
using namespace std;
const long long mxn = 2e5 + 7;
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 7;
int dx[] = {+1, -1, 0, 0};
int dy[] = {0, 0, -1, +1};
int n, t;
char c[mxn];
string a, b;
map<int, int> m;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> t >> a >> b;
  int inv = (n - t), cnt = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i] && inv > 0) {
      c[i] = a[i];
      inv--;
      cnt++;
    } else {
      if (a[i] != 'a' && b[i] != 'a') {
        c[i] = 'a';
      } else if (a[i] != 'b' && b[i] != 'b') {
        c[i] = 'b';
      } else if (a[i] != 'c' && b[i] != 'c') {
        c[i] = 'c';
      }
    }
  }
  inv += inv;
  int keks = 0;
  if (inv > n - cnt) return cout << -1, 0;
  for (int i = 0; i < n; i++) {
    if (a[i] != b[i] && inv > 0) {
      if (keks % 2) {
        c[i] = a[i];
      } else {
        c[i] = b[i];
      }
      inv--;
      keks++;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << c[i];
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 long long mxn = 2e5 + 7;
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 7;
int dx[] = {+1, -1, 0, 0};
int dy[] = {0, 0, -1, +1};
int n, t;
char c[mxn];
string a, b;
map<int, int> m;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> t >> a >> b;
  int inv = (n - t), cnt = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i] && inv > 0) {
      c[i] = a[i];
      inv--;
      cnt++;
    } else {
      if (a[i] != 'a' && b[i] != 'a') {
        c[i] = 'a';
      } else if (a[i] != 'b' && b[i] != 'b') {
        c[i] = 'b';
      } else if (a[i] != 'c' && b[i] != 'c') {
        c[i] = 'c';
      }
    }
  }
  inv += inv;
  int keks = 0;
  if (inv > n - cnt) return cout << -1, 0;
  for (int i = 0; i < n; i++) {
    if (a[i] != b[i] && inv > 0) {
      if (keks % 2) {
        c[i] = a[i];
      } else {
        c[i] = b[i];
      }
      inv--;
      keks++;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << c[i];
  }
  return 0;
}

```