#include <bits/stdc++.h>
using namespace std;
vector<int> v, d;
vector<pair<int, char> > vp;
int main(void) {
  int n, t;
  string a, b, an;
  scanf("%d %d", &n, &t);
  cin >> a;
  cin >> b;
  int s = 0;
  for (int i = 0; i < n; i++)
    if (a[i] == b[i]) {
      s++;
      v.push_back(i);
    } else {
      d.push_back(i);
    }
  if (n - 2 * t > s) {
    printf("-1\n");
    return 0;
  }
  char c;
  if (s >= n - t) {
    for (int i = 0; i < n - t; i++) vp.push_back(make_pair(v[i], a[v[i]]));
    for (int i = n - t; i < s; i++)
      vp.push_back(make_pair(v[i], a[v[i]] == 'z' ? 'a' : a[v[i]] + 1));
    for (int i = 0; i < d.size(); i++) {
      c = a[d[i]] + 1;
      if (c > 'z') c = 'a';
      if (c == b[d[i]]) c++;
      if (c > 'z') c = 'a';
      vp.push_back(make_pair(d[i], c));
    }
    sort(vp.begin(), vp.end());
    for (int i = 0; i < n; i++) an += vp[i].second;
  } else {
    for (int i = 0; i < s; i++) vp.push_back(make_pair(v[i], a[v[i]]));
    for (int i = 0; i < n - t - s; i++) vp.push_back(make_pair(d[i], a[d[i]]));
    for (int i = n - t - s; i < 2 * (n - t - s); i++)
      vp.push_back(make_pair(d[i], b[d[i]]));
    for (int i = 2 * (n - t - s); i < d.size(); i++) {
      c = a[d[i]] + 1;
      if (c > 'z') c = 'a';
      if (c == b[d[i]]) c++;
      if (c > 'z') c = 'a';
      vp.push_back(make_pair(d[i], c));
    }
    sort(vp.begin(), vp.end());
    for (int i = 0; i < n; i++) an += vp[i].second;
  }
  cout << an << endl;
  return 0;
}

### Prompt

Please formulate 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;
vector<int> v, d;
vector<pair<int, char> > vp;
int main(void) {
  int n, t;
  string a, b, an;
  scanf("%d %d", &n, &t);
  cin >> a;
  cin >> b;
  int s = 0;
  for (int i = 0; i < n; i++)
    if (a[i] == b[i]) {
      s++;
      v.push_back(i);
    } else {
      d.push_back(i);
    }
  if (n - 2 * t > s) {
    printf("-1\n");
    return 0;
  }
  char c;
  if (s >= n - t) {
    for (int i = 0; i < n - t; i++) vp.push_back(make_pair(v[i], a[v[i]]));
    for (int i = n - t; i < s; i++)
      vp.push_back(make_pair(v[i], a[v[i]] == 'z' ? 'a' : a[v[i]] + 1));
    for (int i = 0; i < d.size(); i++) {
      c = a[d[i]] + 1;
      if (c > 'z') c = 'a';
      if (c == b[d[i]]) c++;
      if (c > 'z') c = 'a';
      vp.push_back(make_pair(d[i], c));
    }
    sort(vp.begin(), vp.end());
    for (int i = 0; i < n; i++) an += vp[i].second;
  } else {
    for (int i = 0; i < s; i++) vp.push_back(make_pair(v[i], a[v[i]]));
    for (int i = 0; i < n - t - s; i++) vp.push_back(make_pair(d[i], a[d[i]]));
    for (int i = n - t - s; i < 2 * (n - t - s); i++)
      vp.push_back(make_pair(d[i], b[d[i]]));
    for (int i = 2 * (n - t - s); i < d.size(); i++) {
      c = a[d[i]] + 1;
      if (c > 'z') c = 'a';
      if (c == b[d[i]]) c++;
      if (c > 'z') c = 'a';
      vp.push_back(make_pair(d[i], c));
    }
    sort(vp.begin(), vp.end());
    for (int i = 0; i < n; i++) an += vp[i].second;
  }
  cout << an << endl;
  return 0;
}

```