#include <bits/stdc++.h>
using namespace std;
const int ma = 1e5 + 5;
int n, t;
string a, b, ans;
bool vis[ma];
int main() {
  cin >> n >> t;
  cin >> a >> b;
  int same = n - t, cur = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i] && cur < same) {
      cur++;
      vis[i] = true;
    }
  }
  int x[2] = {cur, cur};
  for (int i = 0; i < n; i++) {
    if (vis[i]) {
      ans += a[i];
    } else {
      if (x[0] < same) {
        x[0]++;
        ans += a[i];
      } else {
        if (x[1] < same) {
          x[1]++;
          ans += b[i];
        } else {
          for (int j = 0; j < 26; j++) {
            if (a[i] - 'a' != j && b[i] - 'a' != j) {
              ans += 'a' + j;
              break;
            }
          }
        }
      }
    }
  }
  if (x[0] == same && x[1] == same) {
    cout << ans << endl;
  } else
    cout << -1 << endl;
  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 int ma = 1e5 + 5;
int n, t;
string a, b, ans;
bool vis[ma];
int main() {
  cin >> n >> t;
  cin >> a >> b;
  int same = n - t, cur = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i] && cur < same) {
      cur++;
      vis[i] = true;
    }
  }
  int x[2] = {cur, cur};
  for (int i = 0; i < n; i++) {
    if (vis[i]) {
      ans += a[i];
    } else {
      if (x[0] < same) {
        x[0]++;
        ans += a[i];
      } else {
        if (x[1] < same) {
          x[1]++;
          ans += b[i];
        } else {
          for (int j = 0; j < 26; j++) {
            if (a[i] - 'a' != j && b[i] - 'a' != j) {
              ans += 'a' + j;
              break;
            }
          }
        }
      }
    }
  }
  if (x[0] == same && x[1] == same) {
    cout << ans << endl;
  } else
    cout << -1 << endl;
  return 0;
}

```