#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int inf = 0x3f3f3f3f;
long long infl = 0x3f3f3f3f3f3f3f3fLL;
long double infd = 1.0 / 0.0;
const long long MOD = 1e9 + 7;
const long double pi = 2 * acos(0.0);
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n, c;
  cin >> n >> c;
  string s, t;
  cin >> s >> t;
  if (c == 0) {
    if (s == t) return cout << s, 0;
    return cout << -1, 0;
  }
  int x = n - c;
  vector<int> id(n);
  int diff = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == t[i] && x) {
      x--;
      id[i] = 1;
    }
    if (s[i] != t[i]) diff++;
  }
  if (2 * x > diff) return cout << -1, 0;
  int last = n - 1;
  for (int i = 0; i < n; i++) {
    if (id[i] == -1) {
      cout << t[i];
      continue;
    }
    if (id[i]) {
      cout << s[i];
      continue;
    }
    if (!x) {
      for (int j = 0; j < 3; j++)
        if (s[i] != 'a' + j && t[i] != 'a' + j) {
          cout << char('a' + j);
          break;
        }
      continue;
    }
    cout << s[i];
    x--;
    while (last >= 0) {
      if (id[last]) {
        last--;
        continue;
      }
      id[last] = -1;
      last--;
      break;
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int inf = 0x3f3f3f3f;
long long infl = 0x3f3f3f3f3f3f3f3fLL;
long double infd = 1.0 / 0.0;
const long long MOD = 1e9 + 7;
const long double pi = 2 * acos(0.0);
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n, c;
  cin >> n >> c;
  string s, t;
  cin >> s >> t;
  if (c == 0) {
    if (s == t) return cout << s, 0;
    return cout << -1, 0;
  }
  int x = n - c;
  vector<int> id(n);
  int diff = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == t[i] && x) {
      x--;
      id[i] = 1;
    }
    if (s[i] != t[i]) diff++;
  }
  if (2 * x > diff) return cout << -1, 0;
  int last = n - 1;
  for (int i = 0; i < n; i++) {
    if (id[i] == -1) {
      cout << t[i];
      continue;
    }
    if (id[i]) {
      cout << s[i];
      continue;
    }
    if (!x) {
      for (int j = 0; j < 3; j++)
        if (s[i] != 'a' + j && t[i] != 'a' + j) {
          cout << char('a' + j);
          break;
        }
      continue;
    }
    cout << s[i];
    x--;
    while (last >= 0) {
      if (id[last]) {
        last--;
        continue;
      }
      id[last] = -1;
      last--;
      break;
    }
  }
  return 0;
}

```