#include <bits/stdc++.h>
using namespace std;
void PLAY() {
  cout << fixed << setprecision(10);
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int max(int a, int b) {
  if (a > b) return a;
  return b;
}
int min(int a, int b) {
  if (a < b) return a;
  return b;
}
long long fx4[] = {-1, 1, 0, 0};
long long fy4[] = {0, 0, -1, 1};
long long fx8[] = {-1, -1, -1, 0, 1, 1, 1, 0};
long long fy8[] = {-1, 0, 1, 1, 1, 0, -1, -1};
const double PI = 3.141592653589793;
const long long N = 2e5 + 5;
const long long mod = 1000000007;
long long n, t;
string s1, s2;
set<long long> same;
set<long long> diff;
set<char> alp;
int32_t main() {
  PLAY();
  cin >> n >> t;
  cin >> s1 >> s2;
  for (long long i = 0; i < 26; i++) {
    alp.insert(i + 'a');
  }
  string s3;
  for (long long i = 1; i <= n; i++) s3.push_back('*');
  for (long long i = 0; i < n; i++) {
    if (s1[i] == s2[i])
      same.insert(i);
    else {
      diff.insert(i);
    }
  }
  long long k1 = same.size();
  long long op = n - (k1 + (n - k1) / 2);
  if (t < op) {
    cout << "-1"
         << "\n";
    return 0;
  }
  long long sam = n - t;
  while (1) {
    if (k1 == 0 or sam == 0) break;
    s3[*same.begin()] = s1[*same.begin()];
    same.erase(same.begin());
    k1--;
    sam--;
  }
  while (sam--) {
    auto it = diff.begin();
    s3[*it] = s1[*it];
    it++;
    s3[*it] = s2[*it];
    diff.erase(diff.begin());
    diff.erase(diff.begin());
  }
  for (long long i = 0; i < n; i++) {
    if (s3[i] == '*') {
      char c;
      for (auto k : alp) {
        if (k != s1[i] and k != s2[i]) {
          c = k;
          break;
        }
      }
      s3[i] = c;
    }
  }
  cout << s3 << "\n";
}

### Prompt

Create a solution in CPP for 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;
void PLAY() {
  cout << fixed << setprecision(10);
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int max(int a, int b) {
  if (a > b) return a;
  return b;
}
int min(int a, int b) {
  if (a < b) return a;
  return b;
}
long long fx4[] = {-1, 1, 0, 0};
long long fy4[] = {0, 0, -1, 1};
long long fx8[] = {-1, -1, -1, 0, 1, 1, 1, 0};
long long fy8[] = {-1, 0, 1, 1, 1, 0, -1, -1};
const double PI = 3.141592653589793;
const long long N = 2e5 + 5;
const long long mod = 1000000007;
long long n, t;
string s1, s2;
set<long long> same;
set<long long> diff;
set<char> alp;
int32_t main() {
  PLAY();
  cin >> n >> t;
  cin >> s1 >> s2;
  for (long long i = 0; i < 26; i++) {
    alp.insert(i + 'a');
  }
  string s3;
  for (long long i = 1; i <= n; i++) s3.push_back('*');
  for (long long i = 0; i < n; i++) {
    if (s1[i] == s2[i])
      same.insert(i);
    else {
      diff.insert(i);
    }
  }
  long long k1 = same.size();
  long long op = n - (k1 + (n - k1) / 2);
  if (t < op) {
    cout << "-1"
         << "\n";
    return 0;
  }
  long long sam = n - t;
  while (1) {
    if (k1 == 0 or sam == 0) break;
    s3[*same.begin()] = s1[*same.begin()];
    same.erase(same.begin());
    k1--;
    sam--;
  }
  while (sam--) {
    auto it = diff.begin();
    s3[*it] = s1[*it];
    it++;
    s3[*it] = s2[*it];
    diff.erase(diff.begin());
    diff.erase(diff.begin());
  }
  for (long long i = 0; i < n; i++) {
    if (s3[i] == '*') {
      char c;
      for (auto k : alp) {
        if (k != s1[i] and k != s2[i]) {
          c = k;
          break;
        }
      }
      s3[i] = c;
    }
  }
  cout << s3 << "\n";
}

```