#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
const int M = 1e6 + 7;
const int lim = 2e6;
const int P = 998244353;
const int inf = 0x3f3f3f3f;
int n, t, num;
char a[N], b[N], c[N];
int main() {
  memset(c, 0, sizeof(c));
  scanf("%d%d", &n, &t);
  scanf("%s%s", a + 1, b + 1);
  t = n - t;
  for (int i = 1; i <= n; i++)
    if (a[i] == b[i]) num++;
  if ((n - num) / 2 + num < t) {
    printf("-1\n");
    return 0;
  }
  int k = min(t, num);
  for (int i = 1, j = 1; j <= k && i <= n; i++)
    if (a[i] == b[i]) c[i] = a[i], j++;
  k = t - k;
  for (int i = 1, j = 1; j <= k && i <= n; i++)
    if (a[i] != b[i] && !c[i]) c[i] = a[i], j++;
  for (int i = 1, j = 1; j <= k && 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;
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) printf("%c", c[i]);
  return 0;
}

### Prompt

Generate 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;
const int N = 2e5 + 7;
const int M = 1e6 + 7;
const int lim = 2e6;
const int P = 998244353;
const int inf = 0x3f3f3f3f;
int n, t, num;
char a[N], b[N], c[N];
int main() {
  memset(c, 0, sizeof(c));
  scanf("%d%d", &n, &t);
  scanf("%s%s", a + 1, b + 1);
  t = n - t;
  for (int i = 1; i <= n; i++)
    if (a[i] == b[i]) num++;
  if ((n - num) / 2 + num < t) {
    printf("-1\n");
    return 0;
  }
  int k = min(t, num);
  for (int i = 1, j = 1; j <= k && i <= n; i++)
    if (a[i] == b[i]) c[i] = a[i], j++;
  k = t - k;
  for (int i = 1, j = 1; j <= k && i <= n; i++)
    if (a[i] != b[i] && !c[i]) c[i] = a[i], j++;
  for (int i = 1, j = 1; j <= k && 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;
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) printf("%c", c[i]);
  return 0;
}

```