#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100;
string s1, s2, s3;
vector<int> st;
int main() {
  int n, t;
  cin >> n >> t;
  cin >> s1 >> s2;
  int num1 = 0;
  int pos = 0;
  for (int i = 0; i < s1.size(); i++) {
    if (s1[i] != s2[i]) {
      num1++;
    }
  }
  int k = 0;
  int cd = 0;
  int flag = 0;
  int v = t;
  if (t <= num1 && t >= (num1 + 1) / 2) {
    if (num1 % 2) {
      cd++;
      v--;
    }
    v -= (num1) / 2;
    if (v > 0) {
      for (int k = 1; k <= num1; k++) {
        if (v - k / 2 == 0) {
          cd += k;
          break;
        }
      }
    }
  } else if (t > num1) {
    cd = t;
  } else {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < s1.size(); i++) {
    if (s1[i] == s2[i]) {
      s3 += s1[i];
    } else {
      if (!cd) {
        if (num1 % 2 && pos == i) {
          for (int j = 0; j < 26; j++) {
            if ('a' + j != s1[i] && 'a' + j != s2[i]) {
              s3 += 'a' + j;
              break;
            }
          }
        } else {
          if (!k) {
            s3 += s1[i];
            k = !k;
          } else {
            s3 += s2[i];
            k = !k;
          }
        }
      } else {
        for (int j = 0; j < 26; j++) {
          if ('a' + j != s1[i] && 'a' + j != s2[i]) {
            s3 += 'a' + j;
            break;
          }
        }
        cd--;
      }
    }
  }
  for (int i = 0; i < s1.size() && cd > 0; i++) {
    if (s1[i] == s2[i]) {
      for (int j = 0; j < 26 && cd > 0; j++) {
        if ('a' + j != s1[i]) {
          s3[i] = 'a' + j;
          cd--;
          break;
        }
      }
    }
  }
  if (cd > 0) {
    cout << -1 << endl;
  } else {
    cout << s3 << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 int maxn = 1e5 + 100;
string s1, s2, s3;
vector<int> st;
int main() {
  int n, t;
  cin >> n >> t;
  cin >> s1 >> s2;
  int num1 = 0;
  int pos = 0;
  for (int i = 0; i < s1.size(); i++) {
    if (s1[i] != s2[i]) {
      num1++;
    }
  }
  int k = 0;
  int cd = 0;
  int flag = 0;
  int v = t;
  if (t <= num1 && t >= (num1 + 1) / 2) {
    if (num1 % 2) {
      cd++;
      v--;
    }
    v -= (num1) / 2;
    if (v > 0) {
      for (int k = 1; k <= num1; k++) {
        if (v - k / 2 == 0) {
          cd += k;
          break;
        }
      }
    }
  } else if (t > num1) {
    cd = t;
  } else {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < s1.size(); i++) {
    if (s1[i] == s2[i]) {
      s3 += s1[i];
    } else {
      if (!cd) {
        if (num1 % 2 && pos == i) {
          for (int j = 0; j < 26; j++) {
            if ('a' + j != s1[i] && 'a' + j != s2[i]) {
              s3 += 'a' + j;
              break;
            }
          }
        } else {
          if (!k) {
            s3 += s1[i];
            k = !k;
          } else {
            s3 += s2[i];
            k = !k;
          }
        }
      } else {
        for (int j = 0; j < 26; j++) {
          if ('a' + j != s1[i] && 'a' + j != s2[i]) {
            s3 += 'a' + j;
            break;
          }
        }
        cd--;
      }
    }
  }
  for (int i = 0; i < s1.size() && cd > 0; i++) {
    if (s1[i] == s2[i]) {
      for (int j = 0; j < 26 && cd > 0; j++) {
        if ('a' + j != s1[i]) {
          s3[i] = 'a' + j;
          cd--;
          break;
        }
      }
    }
  }
  if (cd > 0) {
    cout << -1 << endl;
  } else {
    cout << s3 << endl;
  }
  return 0;
}

```