#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
long long power(long long a, long long b);
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long tt = 1;
  while (tt--) {
    long long n, t;
    cin >> n >> t;
    string a, b;
    cin >> a >> b;
    t = n - t;
    string ans(n, '1');
    vector<long long> arr[2];
    for (long long i = 0; i < n; i++) arr[!(a[i] == b[i])].push_back(i);
    long long have = (arr[0].size() + (arr[1].size()) / 2);
    if (have >= t) {
      while (t > 0 && arr[0].size() > 0) {
        long long index = arr[0].back();
        ans[index] = a[index];
        t--;
        arr[0].pop_back();
      }
      while (t > 0 && arr[1].size() > 0) {
        long long index = arr[1].back();
        ans[index] = a[index];
        t--;
        arr[1].pop_back();
        index = arr[1].back();
        ans[index] = b[index];
        arr[1].pop_back();
      }
      while (arr[0].size() > 0) {
        long long index = arr[0].back();
        ans[index] = (a[index] == 'a') ? 'b' : 'a';
        arr[0].pop_back();
      }
      while (arr[1].size() > 0) {
        long long index = arr[1].back();
        for (char c = 'a'; c <= 'z'; c++) {
          if (c != a[index] && c != b[index]) {
            ans[index] = c;
            break;
          }
        }
        arr[1].pop_back();
      }
      cout << ans << "\n";
    } else
      cout << "-1"
           << "\n";
  }
  return 0;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2)
      b--, res = res * a;
    else
      b = b / 2, a *= a;
  }
  return res;
}

### Prompt

Please create a solution in cpp 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;
const long long inf = 1e18;
long long power(long long a, long long b);
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long tt = 1;
  while (tt--) {
    long long n, t;
    cin >> n >> t;
    string a, b;
    cin >> a >> b;
    t = n - t;
    string ans(n, '1');
    vector<long long> arr[2];
    for (long long i = 0; i < n; i++) arr[!(a[i] == b[i])].push_back(i);
    long long have = (arr[0].size() + (arr[1].size()) / 2);
    if (have >= t) {
      while (t > 0 && arr[0].size() > 0) {
        long long index = arr[0].back();
        ans[index] = a[index];
        t--;
        arr[0].pop_back();
      }
      while (t > 0 && arr[1].size() > 0) {
        long long index = arr[1].back();
        ans[index] = a[index];
        t--;
        arr[1].pop_back();
        index = arr[1].back();
        ans[index] = b[index];
        arr[1].pop_back();
      }
      while (arr[0].size() > 0) {
        long long index = arr[0].back();
        ans[index] = (a[index] == 'a') ? 'b' : 'a';
        arr[0].pop_back();
      }
      while (arr[1].size() > 0) {
        long long index = arr[1].back();
        for (char c = 'a'; c <= 'z'; c++) {
          if (c != a[index] && c != b[index]) {
            ans[index] = c;
            break;
          }
        }
        arr[1].pop_back();
      }
      cout << ans << "\n";
    } else
      cout << "-1"
           << "\n";
  }
  return 0;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2)
      b--, res = res * a;
    else
      b = b / 2, a *= a;
  }
  return res;
}

```