#include <bits/stdc++.h>
using namespace std;
const int MX = 1e5 + 5;
char A[MX], B[MX], C[MX];
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    memset(C, 0, sizeof(C));
    scanf("%s%s", A + 1, B + 1);
    m = n - m;
    int sum = 0;
    for (int i = 1; i <= n; i++) {
      if (A[i] == B[i]) sum++;
    }
    if ((n - sum) / 2 + sum < m) {
      printf("-1\n");
      continue;
    }
    int k = min(m, sum);
    for (int i = 1, j = 1; j <= k && i <= n; i++) {
      if (A[i] == B[i]) C[i] = A[i], j++;
    }
    int s = m - k;
    for (int i = 1, j = 1; j <= s && i <= n; i++) {
      if (A[i] != B[i] && !C[i]) C[i] = A[i], j++;
    }
    for (int i = 1, j = 1; j <= s && i <= n; i++) {
      if (A[i] != B[i] && !C[i]) C[i] = B[i], j++;
    }
    for (int i = 1; i <= n; i++) {
      if (!C[i])
        for (int j = 'a'; j <= 'z'; j++) {
          if (A[i] != j && B[i] != j) {
            C[i] = j;
            break;
          }
        }
    }
    printf("%s\n", C + 1);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 MX = 1e5 + 5;
char A[MX], B[MX], C[MX];
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    memset(C, 0, sizeof(C));
    scanf("%s%s", A + 1, B + 1);
    m = n - m;
    int sum = 0;
    for (int i = 1; i <= n; i++) {
      if (A[i] == B[i]) sum++;
    }
    if ((n - sum) / 2 + sum < m) {
      printf("-1\n");
      continue;
    }
    int k = min(m, sum);
    for (int i = 1, j = 1; j <= k && i <= n; i++) {
      if (A[i] == B[i]) C[i] = A[i], j++;
    }
    int s = m - k;
    for (int i = 1, j = 1; j <= s && i <= n; i++) {
      if (A[i] != B[i] && !C[i]) C[i] = A[i], j++;
    }
    for (int i = 1, j = 1; j <= s && i <= n; i++) {
      if (A[i] != B[i] && !C[i]) C[i] = B[i], j++;
    }
    for (int i = 1; i <= n; i++) {
      if (!C[i])
        for (int j = 'a'; j <= 'z'; j++) {
          if (A[i] != j && B[i] != j) {
            C[i] = j;
            break;
          }
        }
    }
    printf("%s\n", C + 1);
  }
  return 0;
}

```