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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
char a[100010], b[100010], c[100010];
int main() {
  int n, k;
  scanf("%d%d\n", &n, &k);
  gets(a + 1);
  gets(b + 1);
  int grupa1 = 0, grupa2 = 0;
  for (int i = 1; i <= n; ++i)
    if (a[i] != b[i])
      ++grupa1;
    else
      ++grupa2;
  for (int x = 0; x <= grupa1 && x <= k; ++x) {
    int rem = k - x;
    int rem_grupa1 = grupa1 - x;
    if (rem_grupa1 % 2) continue;
    rem -= rem_grupa1 / 2;
    if (rem >= 0 && rem <= grupa2) {
      for (int j = 1; j <= n; ++j)
        if (a[j] != b[j]) {
          if (rem_grupa1 > 0) {
            if (rem_grupa1 % 2 == 0)
              c[j] = a[j];
            else
              c[j] = b[j];
            --rem_grupa1;
          } else {
            for (char ch = 'a'; ch <= 'z'; ++ch)
              if (ch != a[j] && ch != b[j]) {
                c[j] = ch;
                break;
              }
          }
        } else {
          if (rem) {
            if (a[j] != 'a')
              c[j] = 'a';
            else
              c[j] = 'b';
            --rem;
          } else
            c[j] = a[j];
        }
      for (int j = 1; j <= n; ++j) printf("%c", c[j]);
      return 0;
    }
  }
  printf("-1");
  return 0;
}