#include <bits/stdc++.h>
using namespace std;
void Yahia74() {
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
}
const int N = 1e5 + 74, M = 1e5 + 74, OO = 0x3f3f3f3f;
char s[N], s1[N], s2[N];
int n, t, cnt, lastIdx;
bool flag = true;
void changeChar(int i) {
  if (s[i] != 'a' && s1[i] != 'a')
    s2[i] = 'a';
  else if (s[i] != 'b' && s1[i] != 'b')
    s2[i] = 'b';
  else if (s[i] != 'c' && s1[i] != 'c')
    s2[i] = 'c';
}
int main() {
  Yahia74();
  cin >> n >> t >> s >> s1;
  for (int i = 0; i < (int)n; i++)
    if (s[i] == s1[i])
      ++cnt;
    else
      lastIdx = i;
  if ((n - cnt + 1) / 2 > t || (!t && cnt != n)) return cout << -1, 0;
  for (int i = 0; i < (int)n; i++) {
    if (s[i] == s1[i])
      s2[i] = s[i];
    else {
      if (flag) {
        flag = false;
        if (i == lastIdx && ((n - cnt) & 1))
          changeChar(i);
        else
          s2[i] = s[i];
      } else {
        flag = true;
        if (i == lastIdx && ((n - cnt) & 1))
          changeChar(i);
        else
          s2[i] = s1[i];
      }
    }
  }
  t -= (n - cnt + 1) / 2;
  flag = true;
  for (int i = 0; i < n && t; ++i) {
    if (((i == lastIdx) && ((n - cnt) & 1))) continue;
    if (s[i] == s1[i])
      continue;
    else {
      if (flag) {
        flag = false;
        changeChar(i);
      } else {
        flag = true;
        --t;
        changeChar(i);
      }
    }
  }
  if (t) {
    for (int i = 0; i < n && t; i++) {
      if (s[i] == s1[i]) {
        changeChar(i);
        --t;
      }
    }
  }
  cout << s2;
  return 0;
}

### 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;
void Yahia74() {
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
}
const int N = 1e5 + 74, M = 1e5 + 74, OO = 0x3f3f3f3f;
char s[N], s1[N], s2[N];
int n, t, cnt, lastIdx;
bool flag = true;
void changeChar(int i) {
  if (s[i] != 'a' && s1[i] != 'a')
    s2[i] = 'a';
  else if (s[i] != 'b' && s1[i] != 'b')
    s2[i] = 'b';
  else if (s[i] != 'c' && s1[i] != 'c')
    s2[i] = 'c';
}
int main() {
  Yahia74();
  cin >> n >> t >> s >> s1;
  for (int i = 0; i < (int)n; i++)
    if (s[i] == s1[i])
      ++cnt;
    else
      lastIdx = i;
  if ((n - cnt + 1) / 2 > t || (!t && cnt != n)) return cout << -1, 0;
  for (int i = 0; i < (int)n; i++) {
    if (s[i] == s1[i])
      s2[i] = s[i];
    else {
      if (flag) {
        flag = false;
        if (i == lastIdx && ((n - cnt) & 1))
          changeChar(i);
        else
          s2[i] = s[i];
      } else {
        flag = true;
        if (i == lastIdx && ((n - cnt) & 1))
          changeChar(i);
        else
          s2[i] = s1[i];
      }
    }
  }
  t -= (n - cnt + 1) / 2;
  flag = true;
  for (int i = 0; i < n && t; ++i) {
    if (((i == lastIdx) && ((n - cnt) & 1))) continue;
    if (s[i] == s1[i])
      continue;
    else {
      if (flag) {
        flag = false;
        changeChar(i);
      } else {
        flag = true;
        --t;
        changeChar(i);
      }
    }
  }
  if (t) {
    for (int i = 0; i < n && t; i++) {
      if (s[i] == s1[i]) {
        changeChar(i);
        --t;
      }
    }
  }
  cout << s2;
  return 0;
}

```