#include <bits/stdc++.h>
using namespace std;
vector<int> diff, sim;
int F(string s1, string s2) {
  int i;
  for (i = 0; i < s1.length(); i++) {
    if (s1[i] - s2[i]) {
      diff.push_back(i);
    } else
      sim.push_back(i);
  }
  return diff.size();
}
char some(char a, char b) {
  char p = 'a';
  for (int i = 0; i < 26; i++) {
    if (p + i != a && p + i != b) return p + i;
  }
}
int main() {
  std::ios::sync_with_stdio(0);
  int n, t;
  cin >> n >> t;
  char Ans[n];
  int i;
  string s1, s2;
  cin >> s1 >> s2;
  int l1 = F(s1, s2), l;
  l = l1;
  if (l % 2 == 0)
    l /= 2;
  else {
    l++;
    l /= 2;
  }
  if (t < l) {
    cout << "-1" << endl;
  } else {
    if (t >= l1) {
      vector<int>::iterator x = diff.begin();
      t -= l1;
      while (x != diff.end()) {
        Ans[*x] = some(s1[*x], s2[*x]);
        x++;
      }
      x = sim.begin();
      while (t) {
        Ans[*x] = some(s1[*x], s2[*x]);
        x++;
        t--;
      }
      while (x != sim.end()) {
        Ans[*x] = s1[*x];
        x++;
      }
      for (int j = 0; j < n; j++) cout << Ans[j];
      cout << endl;
    } else {
      int cnt = l1 - t;
      vector<int>::iterator x = diff.begin();
      while (cnt) {
        Ans[*x] = s1[*x];
        x++;
        cnt--;
      }
      cnt = l1 - t;
      while (cnt) {
        Ans[*x] = s2[*x];
        x++;
        cnt--;
      }
      while (x != diff.end()) {
        Ans[*x] = some(s1[*x], s2[*x]);
        x++;
      }
      x = sim.begin();
      while (x != sim.end()) {
        Ans[*x] = s1[*x];
        x++;
      }
      for (int j = 0; j < n; j++) cout << Ans[j];
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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> diff, sim;
int F(string s1, string s2) {
  int i;
  for (i = 0; i < s1.length(); i++) {
    if (s1[i] - s2[i]) {
      diff.push_back(i);
    } else
      sim.push_back(i);
  }
  return diff.size();
}
char some(char a, char b) {
  char p = 'a';
  for (int i = 0; i < 26; i++) {
    if (p + i != a && p + i != b) return p + i;
  }
}
int main() {
  std::ios::sync_with_stdio(0);
  int n, t;
  cin >> n >> t;
  char Ans[n];
  int i;
  string s1, s2;
  cin >> s1 >> s2;
  int l1 = F(s1, s2), l;
  l = l1;
  if (l % 2 == 0)
    l /= 2;
  else {
    l++;
    l /= 2;
  }
  if (t < l) {
    cout << "-1" << endl;
  } else {
    if (t >= l1) {
      vector<int>::iterator x = diff.begin();
      t -= l1;
      while (x != diff.end()) {
        Ans[*x] = some(s1[*x], s2[*x]);
        x++;
      }
      x = sim.begin();
      while (t) {
        Ans[*x] = some(s1[*x], s2[*x]);
        x++;
        t--;
      }
      while (x != sim.end()) {
        Ans[*x] = s1[*x];
        x++;
      }
      for (int j = 0; j < n; j++) cout << Ans[j];
      cout << endl;
    } else {
      int cnt = l1 - t;
      vector<int>::iterator x = diff.begin();
      while (cnt) {
        Ans[*x] = s1[*x];
        x++;
        cnt--;
      }
      cnt = l1 - t;
      while (cnt) {
        Ans[*x] = s2[*x];
        x++;
        cnt--;
      }
      while (x != diff.end()) {
        Ans[*x] = some(s1[*x], s2[*x]);
        x++;
      }
      x = sim.begin();
      while (x != sim.end()) {
        Ans[*x] = s1[*x];
        x++;
      }
      for (int j = 0; j < n; j++) cout << Ans[j];
      cout << endl;
    }
  }
  return 0;
}

```