#include <bits/stdc++.h>
using namespace std;
int n, t;
char s1[100002], s2[100002], build[100002];
int cnt1, cnt2;
vector<int> dif1, dif2;
int main() {
  scanf("%d %d", &n, &t);
  scanf("%s", s1);
  scanf("%s", s2);
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      cnt2++;
      dif2.push_back(i);
    } else {
      cnt1++;
      dif1.push_back(i);
    }
    build[i] = '-';
  }
  int already = 0;
  if (cnt1 % 2 == 1) {
    int ind = dif1[--cnt1];
    build[ind] = s1[ind];
    do {
      build[ind]++;
      if (build[ind] > 'z') build[ind] = 'a';
    } while (build[ind] == s1[ind] || build[ind] == s2[ind]);
    already = 1;
  }
  while (already + cnt1 / 2 + cnt2 < t && cnt1 >= 2) {
    int ind = dif1[--cnt1];
    build[ind] = s1[ind];
    do {
      build[ind]++;
      if (build[ind] > 'z') build[ind] = 'a';
    } while (build[ind] == s1[ind] || build[ind] == s2[ind]);
    ind = dif1[--cnt1];
    build[ind] = s1[ind];
    do {
      build[ind]++;
      if (build[ind] > 'z') build[ind] = 'a';
    } while (build[ind] == s1[ind] || build[ind] == s2[ind]);
    already += 2;
  }
  for (int i = 0; i < cnt1; i++) {
    if (i & 1) {
      build[dif1[i]] = s1[dif1[i]];
    } else {
      build[dif1[i]] = s2[dif1[i]];
    }
  }
  already += cnt1 / 2;
  if (already > t || already + cnt2 < t) {
    printf("-1\n");
    return 0;
  }
  int lft = t - already;
  for (int i = 0; i < lft; i++) {
    build[dif2[i]] = s1[dif2[i]] + 1;
    if (build[dif2[i]] > 'z') build[dif2[i]] = 'a';
  }
  for (int i = lft; i < cnt2; i++) {
    build[dif2[i]] = s1[dif2[i]];
  }
  for (int i = 0; i < n; i++) {
    if (build[i] == '-') cout << 5 / 0;
    printf("%c", build[i]);
  }
  printf("\n");
}

### Prompt

Your task is to create a cpp solution 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;
int n, t;
char s1[100002], s2[100002], build[100002];
int cnt1, cnt2;
vector<int> dif1, dif2;
int main() {
  scanf("%d %d", &n, &t);
  scanf("%s", s1);
  scanf("%s", s2);
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      cnt2++;
      dif2.push_back(i);
    } else {
      cnt1++;
      dif1.push_back(i);
    }
    build[i] = '-';
  }
  int already = 0;
  if (cnt1 % 2 == 1) {
    int ind = dif1[--cnt1];
    build[ind] = s1[ind];
    do {
      build[ind]++;
      if (build[ind] > 'z') build[ind] = 'a';
    } while (build[ind] == s1[ind] || build[ind] == s2[ind]);
    already = 1;
  }
  while (already + cnt1 / 2 + cnt2 < t && cnt1 >= 2) {
    int ind = dif1[--cnt1];
    build[ind] = s1[ind];
    do {
      build[ind]++;
      if (build[ind] > 'z') build[ind] = 'a';
    } while (build[ind] == s1[ind] || build[ind] == s2[ind]);
    ind = dif1[--cnt1];
    build[ind] = s1[ind];
    do {
      build[ind]++;
      if (build[ind] > 'z') build[ind] = 'a';
    } while (build[ind] == s1[ind] || build[ind] == s2[ind]);
    already += 2;
  }
  for (int i = 0; i < cnt1; i++) {
    if (i & 1) {
      build[dif1[i]] = s1[dif1[i]];
    } else {
      build[dif1[i]] = s2[dif1[i]];
    }
  }
  already += cnt1 / 2;
  if (already > t || already + cnt2 < t) {
    printf("-1\n");
    return 0;
  }
  int lft = t - already;
  for (int i = 0; i < lft; i++) {
    build[dif2[i]] = s1[dif2[i]] + 1;
    if (build[dif2[i]] > 'z') build[dif2[i]] = 'a';
  }
  for (int i = lft; i < cnt2; i++) {
    build[dif2[i]] = s1[dif2[i]];
  }
  for (int i = 0; i < n; i++) {
    if (build[i] == '-') cout << 5 / 0;
    printf("%c", build[i]);
  }
  printf("\n");
}

```