#include <bits/stdc++.h>
using namespace std;
int n, m, k, k1, k2, l, i, j, a, b, d2[100005], d1[100005], d[100005];
string s, s1, s2;
int main() {
  cin >> n >> k;
  cin >> s1 >> s2;
  k = n - k;
  for (i = 0; i < n; i++) {
    d[i] = -1;
    d1[i] = (int)s1[i] - 97;
    d2[i] = (int)s2[i] - 97;
  }
  for (i = 0; i < n; i++) {
    if (k == 0) break;
    if (d1[i] == d2[i]) {
      d[i] = d1[i];
      k--;
    }
  }
  k1 = k;
  k2 = k;
  for (i = 0; i < n; i++) {
    if (d[i] != -1) continue;
    if (k1 == 0) break;
    d[i] = d1[i];
    k1--;
  }
  for (i = 0; i < n; i++) {
    if (d[i] != -1) continue;
    if (k2 == 0) break;
    d[i] = d2[i];
    k2--;
  }
  for (i = 0; i < n; i++) {
    if (d[i] == -1) {
      if (d1[i] == 25 && d2[i] == 0)
        d[i] = 1;
      else if (d1[i] == 0 && d2[i] == 25)
        d[i] = 1;
      else if (d1[i] == 25 || d2[i] == 25)
        d[i] = 0;
      else
        d[i] = max(d1[i], d2[i]) + 1;
    }
  }
  if (k1 != 0 || k2 != 0) {
    cout << -1;
    return 0;
  } else
    for (i = 0; i < n; i++) {
      cout << (char)(d[i] + 97);
    }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int n, m, k, k1, k2, l, i, j, a, b, d2[100005], d1[100005], d[100005];
string s, s1, s2;
int main() {
  cin >> n >> k;
  cin >> s1 >> s2;
  k = n - k;
  for (i = 0; i < n; i++) {
    d[i] = -1;
    d1[i] = (int)s1[i] - 97;
    d2[i] = (int)s2[i] - 97;
  }
  for (i = 0; i < n; i++) {
    if (k == 0) break;
    if (d1[i] == d2[i]) {
      d[i] = d1[i];
      k--;
    }
  }
  k1 = k;
  k2 = k;
  for (i = 0; i < n; i++) {
    if (d[i] != -1) continue;
    if (k1 == 0) break;
    d[i] = d1[i];
    k1--;
  }
  for (i = 0; i < n; i++) {
    if (d[i] != -1) continue;
    if (k2 == 0) break;
    d[i] = d2[i];
    k2--;
  }
  for (i = 0; i < n; i++) {
    if (d[i] == -1) {
      if (d1[i] == 25 && d2[i] == 0)
        d[i] = 1;
      else if (d1[i] == 0 && d2[i] == 25)
        d[i] = 1;
      else if (d1[i] == 25 || d2[i] == 25)
        d[i] = 0;
      else
        d[i] = max(d1[i], d2[i]) + 1;
    }
  }
  if (k1 != 0 || k2 != 0) {
    cout << -1;
    return 0;
  } else
    for (i = 0; i < n; i++) {
      cout << (char)(d[i] + 97);
    }
}

```