#include <bits/stdc++.h>
using namespace std;
char s1[100005], s2[100005];
char s[100005];
bool same[100005];
void diff(int i) {
  char ch;
  for (ch = 'a'; ch <= 'z'; ch++) {
    if (ch != s1[i] && ch != s2[i]) {
      s[i] = ch;
      return;
    }
  }
}
vector<int> slis;
vector<int> dlis;
int main() {
  int n, t, T, i, j, J, I, k, m;
  char ch;
  cin >> n >> k;
  k = n - k;
  for (i = 0; i < n; i++) cin >> s1[i];
  int sum = 0;
  for (i = 0; i < n; i++) {
    cin >> s2[i];
    same[i] = (s1[i] == s2[i]);
    sum += same[i];
    if (same[i])
      slis.push_back(i);
    else
      dlis.push_back(i);
    diff(i);
  }
  int dif = n - sum;
  if ((sum + (dif / 2)) < k) {
    cout << "-1";
    return 0;
  }
  i = 0;
  while (i < k) {
    if (slis.size() == 0) break;
    s[slis.back()] = s1[slis.back()];
    slis.pop_back();
    i++;
  }
  while (i < k) {
    s[dlis.back()] = s1[dlis.back()];
    dlis.pop_back();
    s[dlis.back()] = s2[dlis.back()];
    dlis.pop_back();
    i++;
  }
  for (i = 0; i < n; i++) cout << s[i];
  return 0;
}

### Prompt

In cpp, your task is to solve 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 s1[100005], s2[100005];
char s[100005];
bool same[100005];
void diff(int i) {
  char ch;
  for (ch = 'a'; ch <= 'z'; ch++) {
    if (ch != s1[i] && ch != s2[i]) {
      s[i] = ch;
      return;
    }
  }
}
vector<int> slis;
vector<int> dlis;
int main() {
  int n, t, T, i, j, J, I, k, m;
  char ch;
  cin >> n >> k;
  k = n - k;
  for (i = 0; i < n; i++) cin >> s1[i];
  int sum = 0;
  for (i = 0; i < n; i++) {
    cin >> s2[i];
    same[i] = (s1[i] == s2[i]);
    sum += same[i];
    if (same[i])
      slis.push_back(i);
    else
      dlis.push_back(i);
    diff(i);
  }
  int dif = n - sum;
  if ((sum + (dif / 2)) < k) {
    cout << "-1";
    return 0;
  }
  i = 0;
  while (i < k) {
    if (slis.size() == 0) break;
    s[slis.back()] = s1[slis.back()];
    slis.pop_back();
    i++;
  }
  while (i < k) {
    s[dlis.back()] = s1[dlis.back()];
    dlis.pop_back();
    s[dlis.back()] = s2[dlis.back()];
    dlis.pop_back();
    i++;
  }
  for (i = 0; i < n; i++) cout << s[i];
  return 0;
}

```